DP Maths AA HL - Geometry and Trigonometry - Coordinate Geometry

DP Maths AA HL - Geometry and Trigonometry - Coordinate Geometry - Full notes

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DP Maths AA HL | Coordinate Geometry

Coordinate Geometry

This topic contains 4 notes:

3.12 Vectors and Scalar Product
3.14 Vector Equations of Lines
3.16 Vector Product and Planes
1.16 Systems of Linear Equations

DP MATHEMATICS AA · TOPIC 3: GEOMETRY AND TRIGONOMETRY

3.12

Vectors and the Scalar Product

Full notes for HL · Syllabus AHL 3.12, 3.13

Contents

1	Vectors and scalars
2	Representing vectors
3	Vector algebra
4	Magnitude and unit vectors
5	Position vectors and displacement
6	Parallel vectors and collinear points
7	The scalar (dot) product
8	Angles between vectors and perpendicular vectors
9	Geometric proofs with vectors
10	Vectors in geometric figures
11	Unit vectors and vectors of given length
12	The scalar product: algebraic properties
13	Angles in triangles and solids using vectors
14	Perpendicular and parallel vectors with unknowns
15	Projections and components
16	Vectors in kinematics
17	Proof with vectors
18	Mixed review: worked problems
19	Additional exam-style practice
20	Coordinates and vectors in 3D solids
21	Vectors and the cosine rule
22	Exam-style questions with markscheme
23	Common misconceptions and links
24	Check your understanding
25	Summary and topic map
26	Answers to practice exercises

These are full teaching notes. Read each section in order, try the Think about it question before you read on, and work through every worked example before attempting the Quick check.

Before you start

Rate yourself on each statement now, and again after you've worked through the notes. Shade R (not yet), A (getting there) or G (confident).

I can…	R	A	G
distinguish vectors from scalars and represent vectors in column and i, j, k form
add, subtract and multiply vectors by scalars, algebraically and geometrically
find the magnitude of a vector and a unit vector in a given direction
use position vectors and find displacement vectors between points
identify parallel vectors and show that points are collinear
calculate the scalar product and use it to find angles
test vectors for perpendicularity
use vectors to prove geometric results

This booklet covers Additional Higher Level (AHL) content (AHL 3.12, 3.13). It is studied by HL students only.

Guiding questions

How can quantities with both size and direction be represented and combined? How does the scalar product connect algebra to angles?

How this topic is examined

Vectors are tested on both papers, often as the start of a longer lines-and-planes question (3.14, 3.16). Typical tasks: find magnitudes and unit vectors, find angles with the scalar product, show vectors are perpendicular or points collinear.

Distinguish clearly between position vectors (points) and direction vectors. Underline vectors in handwriting.

Command term	What the examiner wants
Write down	an answer with little or no working; it should be obtainable by inspection
Find / Calculate / Determine	obtain the answer showing relevant working
Solve	find the values of the unknowns, using algebra or technology as allowed
Show that	obtain the given result; every step must be shown and the result must not be assumed
Hence / Hence or otherwise	use the previous result (or another valid method)
Sketch	a diagram showing the general shape and key features, labelled
Prove	a formal, rigorous argument from accepted facts
Justify / Explain	give valid reasons or evidence

How to use these notes

Each section teaches one part of the syllabus. You'll see these boxes throughout; each has one job:

Box	What it's for
Think about it	a question to get you thinking before a section
Why it works	a derivation or proof of a key result, so you understand where formulas come from
Key point	a result or idea you must not forget
Common mistake	an error examiners see every year, and how to avoid it
GDC tip	how to use your graphic display calculator efficiently on Paper 2
In the real world	where the mathematics is used outside the classroom
Worked example	a full solution with IB-style marks (M1 method, A1 answer, R1 reasoning, AG answer given); examples are graded basic, standard and exam level
Practice exercise	graded questions for each section; full answers are at the end of the booklet
Quick check	short questions to check the key ideas

1 Vectors and scalars

Size and direction.

Think about it

A plane flies at 500 km/h, but a wind blows at 80 km/h from the west. Why is 'speed 500 km/h' not enough information to predict where the plane ends up?

A scalar has magnitude only: mass, temperature, speed, distance. A vector has magnitude and direction: displacement, velocity, force, acceleration. Two vectors are equal if they have the same magnitude and direction, regardless of where they are drawn.

2 Representing vectors

Arrows, columns and base vectors.

A vector can be drawn as a directed line segment (an arrow), written as AB with an arrow above it for the vector from A to B, or written in bold (v) or underlined (v with a line under it) in handwriting.

In component form, a vector in 3D is written as a column vector or using the base vectors i, j, k (unit vectors along the x, y and z axes):

3 Vector algebra

Combine component by component.

Addition: add corresponding components; geometrically, place the vectors head to tail (triangle rule) or use the parallelogram rule. Subtraction: a − b = a + (−b), where −b has the same magnitude but opposite direction. Scalar multiplication: ka multiplies every component by k, scaling the length by |k| and reversing the direction if k < 0. The zero vector 0 has all components zero.

Figure 1. The parallelogram rule: a + b is the diagonal.

WORKED EXAMPLE 1 Vector arithmetic (basic)

Given a = 2i − j + 3k and b = −i + 4j + k, find 3a − 2b.

Solution

3a = 6i − 3j + 9k; 2b = −2i + 8j + 2kM1

3a − 2b = 8i − 11j + 7kA1

Practice exercise 1

1. Given p = (1, 3, −2) and q = (4, 0, 5) as column vectors, find 2p + q and p − 3q.

2. Find x and y if x(2, 1) + y(1, −1) = (7, 2).

4 Magnitude and unit vectors

Length by Pythagoras.

A unit vector has magnitude 1. To find a unit vector in the direction of v, divide v by its magnitude. To find a vector of length L in the direction of v, multiply the unit vector by L.

WORKED EXAMPLE 2 Unit vector and scaling (standard)

Find a unit vector in the direction of v = 2i − j + 2k, and a vector of length 12 in the opposite direction.

Solution

|v| = √(4 + 1 + 4) = 3A1

Unit vector (1/3)(2i − j + 2k)A1

Length 12, opposite: −4(2i − j + 2k) = −8i + 4j − 8kA1

Practice exercise 2

1. Find |3i + 4j − 12k|.

2. Find k > 0 if |(k, 2k, 2)| = 7.

3. Find the vector of magnitude 10 in the direction of (6, −8, 0).

5 Position vectors and displacement

Points as vectors from the origin.

The position vector of a point A is the vector OA from the origin to A, often written a. The displacement from A to B is AB = b − a ('end minus start'). The midpoint M of AB has position vector m = ½(a + b). The distance AB is |b − a|.

WORKED EXAMPLE 3 Displacement and distance (standard)

A(1, 2, −1) and B(4, −2, 11). Find AB and its length, and the position vector of the point P dividing AB in the ratio 1 : 2.

Solution

AB = b − a = (3, −4, 12)A1

|AB| = √(9 + 16 + 144) = 13A1

p = a + ⅓AB = (1, 2, −1) + (1, −4/3, 4) = (2, 2/3, 3)M1 A1

Practice exercise 3

1. Find the midpoint of A(3, −1, 4) and B(5, 7, −2).

2. ABCD is a parallelogram with A(1, 0, 2), B(4, 1, 3), C(6, 5, 1). Find D.

6 Parallel vectors and collinear points

Scalar multiples.

Two non-zero vectors are parallel if one is a scalar multiple of the other: b = ka. Points A, B and C are collinear (lie on one straight line) if AB and AC (or AB and BC) are parallel and share a point.

WORKED EXAMPLE 4 Collinearity (exam level)

Show that A(1, 2, 3), B(3, 5, 4) and C(7, 11, 6) are collinear.

Solution

AB = (2, 3, 1); AC = (6, 9, 3)A1

AC = 3AB, so AB is parallel to ACR1

They share point A, so A, B, C are collinearR1

Practice exercise 4

1. Find p and q if (2, p, 6) is parallel to (3, 6, q).

2. Determine whether (1, 1, 1), (2, 3, 5) and (4, 7, 12) are collinear.

7 The scalar (dot) product

Multiply and add the components.

The scalar product of two vectors is a number (a scalar), not a vector. It equals the product of the magnitudes times the cosine of the angle θ between them (placed tail to tail). Properties: v·w = w·v; v·(w + u) = v·w + v·u; v·v = |v|2; i·i = 1, i·j = 0.

Why it works

Draw v and w tail to tail with angle θ, so the third side of the triangle is v − w. The cosine rule gives |v − w|2 = |v|2 + |w|2 − 2|v||w| cos θ. Expanding the left side in components gives |v|2 + |w|2 − 2(v1w1 + v2w2 + v3w3). Comparing, v1w1 + v2w2 + v3w3 = |v||w| cos θ.

WORKED EXAMPLE 5 Scalar product (basic)

Find (2i − 3j + k)·(4i + j − 2k).

Solution

2(4) + (−3)(1) + 1(−2)M1

= 3A1

8 Angles between vectors and perpendicular vectors

cos θ = v·w/(|v||w|).

Rearranging gives the angle between two vectors. Two non-zero vectors are perpendicular if and only if their scalar product is zero (cos 90° = 0). If v·w > 0 the angle is acute; if v·w < 0 it is obtuse.

WORKED EXAMPLE 6 Angle between vectors (standard)

Find the angle between a = (1, 2, 2) and b = (3, 0, −4).

Solution

a·b = 3 + 0 − 8 = −5; |a| = 3, |b| = 5A1

cos θ = −5/15 = −1/3M1

θ = 109° (109.5°)A1

WORKED EXAMPLE 7 Finding a perpendicular vector (standard)

Find t such that (2, t, −1) is perpendicular to (3, −2, 4).

Solution

6 − 2t − 4 = 0M1

t = 1A1

WORKED EXAMPLE 8 Angle in a triangle (exam level)

Find angle ABC for A(1, 0, 2), B(2, 1, 0) and C(0, 3, 1).

Solution

Use BA and BC (both from B): BA = (−1, −1, 2), BC = (−2, 2, 1)M1

BA·BC = 2 − 2 + 2 = 2; |BA| = √6, |BC| = 3A1

cos B = 2/(3√6), B = 74.2°A1

Common mistake

Using AB and BC to find the angle at B. Both vectors must start at B (BA and BC); otherwise you get the supplementary angle.

Practice exercise 5

1. Find the angle between (1, 1, 0) and (0, 1, 1).

2. Find p if (p, 2, 3) and (2, p, −4) are perpendicular.

3. Show that the triangle A(1, 1, 1), B(2, 3, 2), C(3, 0, 1) has a right angle at A.

4. Find the angle between the diagonals of a cube (use (1, 1, 1) and (1, 1, −1)).

9 Geometric proofs with vectors

Algebra proves geometry.

WORKED EXAMPLE 9 Diagonals of a rhombus (exam level)

Prove that the diagonals of a rhombus are perpendicular.

Solution

Let the sides be a and b with |a| = |b|; diagonals a + b and a − bM1

(a + b)·(a − b) = |a|2 − |b|2A1

= 0 since |a| = |b|, so the diagonals are perpendicularR1

Practice exercise 6

1. Prove that the angle in a semicircle is a right angle using vectors (centre O, radius vector a, endpoints ±a, point P with position vector p, |p| = |a|).

2. Show that the midpoints of the sides of any quadrilateral form a parallelogram.

10 Vectors in geometric figures

Paths through a diagram.

In a figure, express vectors as sums of known vectors along a path: if OA = a and OB = b, then AB = b − a, and the midpoint M of AB has OM = ½(a + b). A point dividing AB in the ratio m : n has position vector (na + mb)/(m + n).

WORKED EXAMPLE 10 Parallelogram (standard)

OABC is a parallelogram with OA = a and OC = c. M is the midpoint of AB. Express OM and CM in terms of a and c.

Solution

OB = a + c; OM = a + ½cA1 A1

CM = OM − OC = a − ½cA1

WORKED EXAMPLE 11 Ratio (standard)

P divides AB in the ratio 2 : 3. Find OP in terms of a and b.

Solution

OP = a + (2/5)(b − a)M1

= (3a + 2b)/5A1

Practice exercise 7

1. In triangle OAB, M is the midpoint of OA and N the midpoint of OB. Show that MN is parallel to AB and half its length.

2. Find the point dividing A(1, 2, 3) to B(7, −4, 9) in the ratio 1 : 2.

11 Unit vectors and vectors of given length

Scaling directions.

The unit vector in the direction of v is v/|v|. A vector of length L in that direction is Lv/|v|. Use this to place a point a given distance along a direction.

WORKED EXAMPLE 12 Given length (standard)

Find a vector of length 15 in the direction of (2, −2, 1).

Solution

|v| = 3A1

(10, −10, 5)A1

Practice exercise 8

1. Find the unit vector in the direction of 6i − 2j + 3k.

2. A drone moves 26 m in the direction (12, 5, 0) from (1, 1, 10). Find its new position.

12 The scalar product: algebraic properties

Expanding brackets.

The scalar product distributes over addition: a·(b + c) = a·b + a·c, and a·a = |a|2. This lets you expand expressions like |a + b|2 = |a|2 + 2a·b + |b|2.

WORKED EXAMPLE 13 Expanding (exam level)

|a| = 3, |b| = 5 and a·b = 6. Find |a + b| and |a − 2b|.

Solution

|a + b|2 = 9 + 12 + 25 = 46: √46M1 A1

|a − 2b|2 = 9 − 24 + 100 = 85: √85A1

Practice exercise 9

1. |a| = 4, |b| = 3, angle between them 60°. Find |a − b|.

2. Show that (a + b)·(a − b) = |a|2 − |b|2.

13 Angles in triangles and solids using vectors

Vectors from the vertex.

To find angle ABC, use the vectors BA and BC (both starting at B). In 3D solids, set up coordinates for the vertices first.

WORKED EXAMPLE 14 Cube (exam level)

A cube has vertices O(0, 0, 0) and G(2, 2, 2). Find the angle between the diagonal OG and the edge along the x-axis.

Solution

OG = (2, 2, 2), edge (1, 0, 0)A1

cos θ = 2/(2√3)M1

θ = 54.7°A1

Practice exercise 10

1. Find angle BAC for A(1, 0, 0), B(0, 2, 0), C(0, 0, 3).

2. Find the angles of the triangle with vertices (0, 0, 0), (3, 0, 0), (0, 4, 0).

14 Perpendicular and parallel vectors with unknowns

Equations from conditions.

Perpendicular: a·b = 0 gives one equation. Parallel: b = ka gives three component equations, which must be consistent.

WORKED EXAMPLE 15 Both conditions (exam level)

Find p and q such that (p, 2, q) is parallel to (3, −1, 4), and find t such that (t, 1, 2) is perpendicular to (3, −1, 4).

Solution

k = −2: p = −6, q = −8A1 A1

3t − 1 + 8 = 0: t = −7/3A1

Practice exercise 11

1. Find the values of x for which (x, 2x, 3) and (x, −1, 1) are perpendicular.

15 Projections and components

How much of a lies along b.

The scalar projection of a onto b is a·b/|b|: the length of the 'shadow' of a along b. The vector projection is (a·b/|b|2)b. Projections give the component of a force in a direction, or the foot of a perpendicular.

WORKED EXAMPLE 16 Projection (standard)

Find the scalar projection of a = (3, 4, 0) onto b = (1, 0, 0) and onto c = (1, 1, 1).

Solution

onto b: 3A1

onto c: 7/√3 = 4.04A1

Practice exercise 12

1. Find the component of the force (10, 5, −2) N in the direction (2, 1, 2).

2. Find the vector projection of (2, 3) onto (1, 1).

16 Vectors in kinematics

Velocity and displacement.

An object moving with constant velocity v from position a has position r = a + tv after time t. Speed is |v|. Relative position of B from A is rB − rA.

WORKED EXAMPLE 17 Ship (standard)

A ship starts at (2, 3) km with velocity (4, −1) km/h. Find its position after 3 hours and its speed.

Solution

(2, 3) + 3(4, −1) = (14, 0)A1

Speed √17 = 4.12 km/hA1

Practice exercise 13

1. Two particles start at (0, 0) and (10, 0) with velocities (2, 1) and (−1, 2). Find their distance apart after 2 s.

17 Proof with vectors

Geometric results.

Vectors prove geometric facts concisely: diagonals of a parallelogram bisect each other; the line joining midpoints of two sides of a triangle is parallel to the third side; the angle in a semicircle is a right angle (using the scalar product).

WORKED EXAMPLE 18 Parallelogram diagonals (exam level)

Prove that the diagonals of a parallelogram bisect each other.

Solution

With OA = a, OC = c, OB = a + c: midpoint of OB is ½(a + c)A1

Midpoint of AC is ½(a + c)A1

Same point, so they bisect each otherR1

Practice exercise 14

1. Prove that the medians of a triangle meet at (a + b + c)/3.

18 Mixed review: worked problems

Bringing the ideas together.

WORKED EXAMPLE 19 Mixed problem (exam level)

A(1, 2, 1), B(3, 1, 4), C(4, 5, 2). (a) Find AB and AC. (b) Find angle BAC. (c) Find the area of the triangle.

Solution

(a) (2, −1, 3) and (3, 3, 1)A1

(b) dot product 6; |AB| = √14, |AC| = √19: 68.4°M1 A1

Practice exercise 15

1. Find D such that ABCD is a parallelogram for A, B, C above.

19 Additional exam-style practice

Paper 1 questions.

WORKED EXAMPLE 20 Paper 1 question (exam level)

Vectors a = (2, 1, −2) and b = (1, −1, k). (a) Find |a|. (b) Find k if a and b are perpendicular. (c) For k = 1, find the angle between a and b.

Solution

(a) 3A1

(b) 2 − 1 − 2k = 0: k = 1/2A1

Practice exercise 16

1. Find a vector perpendicular to both (1, 0, 1) and (0, 1, 1).

20 Coordinates and vectors in 3D solids

Setting up the frame.

Place one vertex at the origin and align edges with the axes. Then every vertex has simple coordinates, and lengths and angles follow from vector methods.

WORKED EXAMPLE 21 Pyramid (exam level)

A pyramid has square base with vertices (0, 0, 0), (4, 0, 0), (4, 4, 0), (0, 4, 0) and apex (2, 2, 6). Find the angle between a slant edge and the base.

Solution

Edge (2, 2, 6); its projection (2, 2, 0)A1

tan θ = 6/(2√2)M1

θ = 64.8°A1

Practice exercise 17

1. Find the length of a slant edge of the pyramid.

21 Vectors and the cosine rule

Two views of the same result.

The scalar product formula a·b = |a||b| cos θ is equivalent to the cosine rule: |a − b|2 = |a|2 + |b|2 − 2|a||b| cos θ. Either can be used to find angles in triangles given coordinates.

WORKED EXAMPLE 22 Two methods (standard)

Find angle P for P(0, 0), Q(4, 0), R(1, 3) using vectors.

Solution

PQ·PR = 4; |PQ| = 4, |PR| = √10A1

cos P = 1/√10: 71.6°A1

22 Exam-style questions with markscheme

Attempt each question under timed conditions before reading the markscheme.

Question 1 (Paper 1 style) (about 15 minutes)

Points A(2, −1, 3), B(4, 1, 2) and C(1, 3, k).

(a) Find AB. [1]

(b) Find the value of k for which AB is perpendicular to AC. [3]

Markscheme

Part	Answer	Marks
(a)	(2, 2, −1)	A1
(b)	AC = (−1, 4, k − 3)	A1
	−2 + 8 − k + 3 = 0	M1
	k = 9	A1
(c)	AC = (−1, 4, −2); AB·AC = 8	A1
	\|AB\| = 3, \|AC\| = √21	A1
	cos A = 8/(3√21)	M1
	54.4°	A1

Examiner's comments. Using vectors from A in both cases was essential.

Question 2 (Paper 2 style) (about 10 minutes)

A drone flies from O with velocity v = (3, 4, 1.2) m s−1.

(a) Find the speed of the drone. [2]

(b) Find the angle between its path and the horizontal. [3]

Markscheme

Part	Answer	Marks
(a)	√(9 + 16 + 1.44)	M1
	5.14 m s−1	A1
(b)	horizontal speed 5	A1
	tan θ = 1.2/5	M1
	13.5°	A1
(c)	(75, 100, 30)	A1

Examiner's comments. The horizontal component is (3, 4, 0), of magnitude 5.

23 Common misconceptions and links

Students often think…	Actually…
the scalar product is a vector	it is a number
AB = a − b	AB = b − a
vectors are equal only if they start at the same point	same magnitude and direction suffices
perpendicular means v·w = 1	v·w = 0

How 3.12 connects to the rest of the course

Topic	Connection
3.14 Lines	direction vectors
3.16 Planes	normal vectors and the vector product
3.1 Triangles	cosine rule
1.12 Complex numbers	the Argand diagram as vectors

24 Check your understanding

Section A: multiple choice

1. |(2, −1, 2)| =

A. 3 B. 5 C. √5 D. 9

2. (1, 2, 3)·(3, 0, −1) =

A. 6 B. −3 C. 0 D. (3, 0, −3)

3. If v·w < 0 the angle is

A. acute B. obtuse C. right D. zero

4. AB from A(1, 1) to B(4, 5) is

A. (5, 6) B. (−3, −4) C. (4, 5) D. (3, 4)

5. Parallel vectors satisfy

A. a·b = 0 B. b = ka C. |a| = |b| D. a + b = 0 only

Section B: structured

6. Find a unit vector perpendicular to both i + j and j + k. [3]

7. Find the angle between 2i − j + k and i + 3j − 2k. [3]

Answers

1. A 2. C 3. B 4. D 5. B

6. (1, −1, 1)/√3 (check: dot products zero).

7. dot product −3; cos θ = −3/(√6 √14): 109°.

25 Summary and topic map

Idea	Remember
Vectors	magnitude and direction; column or i, j, k form
Algebra	add components; scale; triangle and parallelogram rules
Magnitude	√(v12 + v22 + v32); unit vector v/\|v\|
Position vectors	AB = b − a; midpoint ½(a + b)
Parallel	b = ka; collinear points share a point
Scalar product	v·w = Σviwi = \|v\|\|w\| cos θ
Perpendicular	v·w = 0

Before the exam, make sure you…

•use 'end minus start'
•take both vectors from the vertex
•check perpendicularity with the dot product
•state collinearity reasons

Vectors and the Scalar Product on one page

Cover this page with a sheet of paper and try to rebuild it from memory. Whatever you can't recall is what to revise next.

Figure 2. Topic map for Vectors and the Scalar Product.

26 Answers to practice exercises

Check your working against these answers. Full methods are expected in the exam.

Practice exercise 1

1. (6, 6, 1); (−11, 3, −17)

2. 2x + y = 7, x − y = 2: x = 3, y = 1

Practice exercise 2

1. 13

2. 5k2 = 45: k = 3

3. (6, −8, 0)

Practice exercise 3

1. (4, 3, 1)

2. d = a + c − b = (3, 4, 0)

Practice exercise 4

1. k = 3/2: p = 4, q = 9

2. AB = (1, 2, 4), AC = (3, 6, 11): not multiples, so not collinear

Practice exercise 5

1. 60°

2. 4p = 12: p = 3

3. AB = (1, 2, 1), AC = (2, −1, 0): AB·AC = 0

4. cos θ = 1/3: 70.5°

Practice exercise 6

1. (p − a)·(p + a) = |p|2 − |a|2 = 0

2. both pairs of opposite midpoint vectors equal ½ of a diagonal

Practice exercise 7

1. MN = ½(b − a) = ½AB

2. (3, 0, 5)

Practice exercise 8

1. (6, −2, 3)/7

2. (25, 11, 10)

Practice exercise 9

1. √13

2. expand; a·b terms cancel

Practice exercise 10

1. cos A = 1/√(5 × 10): 81.9°

2. 90°, 53.1°, 36.9°

Practice exercise 11

1. x2 − 2x + 3 = 0 has no real solutions: none

Practice exercise 12

1. (20 + 5 − 4)/3 = 7 N

2. (2.5, 2.5)

Practice exercise 13

1. positions (4, 2) and (8, 4): √20 = 4.47

Practice exercise 14

1. each median's point 2/3 of the way from the vertex is (a + b + c)/3

Practice exercise 15

1. D = A + C − B = (2, 6, −1)

Practice exercise 16

1. (−1, −1, 1)

Practice exercise 17

1. √44 = 6.63

DP MATHEMATICS AA · TOPIC 3: GEOMETRY AND TRIGONOMETRY

3.14

Vector Equations of Lines

Full notes for HL · Syllabus AHL 3.14, 3.15

Contents

1	The vector equation of a line
2	Parametric and Cartesian forms
3	Points on a line
4	The angle between two lines
5	Intersecting, parallel and skew lines
6	Shortest distance from a point to a line
7	Lines in kinematics
8	Closest approach
9	Lines in two dimensions with vectors
10	Choosing direction and position vectors
11	Intersection problems with unknowns
12	Angle between lines: further practice
13	Foot of the perpendicular and reflections
14	Shortest distance between skew lines
15	Kinematics: collision and closest approach
16	Lines and geometric shapes
17	Mixed review: worked problems
18	Additional exam-style practice
19	Vector equations with technology
20	Interpreting parameters
21	Exam-style questions with markscheme
22	Common misconceptions and links
23	Check your understanding
24	Summary and topic map
25	Answers to practice exercises

These are full teaching notes. Read each section in order, try the Think about it question before you read on, and work through every worked example before attempting the Quick check.

Before you start

Rate yourself on each statement now, and again after you've worked through the notes. Shade R (not yet), A (getting there) or G (confident).

I can…	R	A	G
write the vector equation of a line through a point with a given direction, or through two points
convert between vector, parametric and Cartesian forms
find the acute angle between two lines
determine whether two lines intersect, are parallel or are skew, and find points of intersection
find the shortest distance from a point to a line
use r = a + tv to model motion with constant velocity
find when two moving objects are closest

This booklet covers Additional Higher Level (AHL) content (AHL 3.14, 3.15). It is studied by HL students only.

Guiding questions

How can a straight line in three dimensions be described with vectors? How can we tell whether two lines in space meet?

How this topic is examined

Lines in 3D are a frequent long question on Papers 1 and 2: intersection or skewness, angles between lines, the foot of a perpendicular, and kinematics problems with ships, aircraft or drones.

Use different parameters (λ and μ) for different lines. Show the equations you solve and check all three components.

Command term	What the examiner wants
Write down	an answer with little or no working; it should be obtainable by inspection
Find / Calculate / Determine	obtain the answer showing relevant working
Solve	find the values of the unknowns, using algebra or technology as allowed
Show that	obtain the given result; every step must be shown and the result must not be assumed
Hence / Hence or otherwise	use the previous result (or another valid method)
Sketch	a diagram showing the general shape and key features, labelled
Prove	a formal, rigorous argument from accepted facts
Justify / Explain	give valid reasons or evidence

How to use these notes

Each section teaches one part of the syllabus. You'll see these boxes throughout; each has one job:

Box	What it's for
Think about it	a question to get you thinking before a section
Why it works	a derivation or proof of a key result, so you understand where formulas come from
Key point	a result or idea you must not forget
Common mistake	an error examiners see every year, and how to avoid it
GDC tip	how to use your graphic display calculator efficiently on Paper 2
In the real world	where the mathematics is used outside the classroom
Worked example	a full solution with IB-style marks (M1 method, A1 answer, R1 reasoning, AG answer given); examples are graded basic, standard and exam level
Practice exercise	graded questions for each section; full answers are at the end of the booklet
Quick check	short questions to check the key ideas

1 The vector equation of a line

A point plus multiples of a direction.

Think about it

In 2D, a line is fixed by a point and a gradient. In 3D there is no single 'gradient'. What replaces it?

A line through the point with position vector a, in the direction of the vector b, contains every point whose position vector is:

b is the direction vector and λ is a parameter: each value of λ gives one point on the line. For a line through two points A and B, use a as the point and b − a as the direction. The same line has many vector equations: any point on it and any non-zero multiple of the direction vector will do.

Figure 1. The line r = a + λb passes through the point with position vector a in the direction b.

WORKED EXAMPLE 1 Line through two points (basic)

Find a vector equation of the line through A(1, −2, 3) and B(4, 0, −1).

Solution

Direction AB = (3, 2, −4)A1

r = (1, −2, 3) + λ(3, 2, −4)A1

Practice exercise 1

1. Write the vector equation of the line through (2, 0, 5) parallel to i − 3k.

2. Write the vector equation of the line through (1, 1, 1) and (3, −1, 4).

3. Write the vector equation of the z-axis.

2 Parametric and Cartesian forms

Three equivalent descriptions.

Writing r = (x, y, z) and separating components gives the parametric equations: x = a1 + λb1, y = a2 + λb2, z = a3 + λb3. Solving each for λ and equating gives the Cartesian form:

If a component of the direction vector is zero, that coordinate is constant: for b = (2, 0, 5), write (x − a1)/2 = (z − a3)/5 with y = a2.

WORKED EXAMPLE 2 Converting forms (standard)

Write the line (x − 3)/2 = (y + 1)/(−4) = z/5 in vector form.

Solution

Point (3, −1, 0); direction (2, −4, 5)A1

r = (3, −1, 0) + λ(2, −4, 5)A1

Practice exercise 2

1. Write r = (1, 4, −2) + λ(3, −1, 2) in Cartesian form.

2. Write (2x − 1)/4 = (y + 3)/2 = 3 − z in vector form.

3 Points on a line

Find λ from one component, check the others.

WORKED EXAMPLE 3 Testing a point (standard)

Determine whether P(7, 2, −5) lies on r = (1, −2, 3) + λ(3, 2, −4).

Solution

x: 1 + 3λ = 7 gives λ = 2M1

y: −2 + 4 = 2 ✓; z: 3 − 8 = −5 ✓A1

P lies on the lineR1

Practice exercise 3

1. Find a and b if (a, 5, b) lies on r = (2, 1, 0) + λ(1, 2, −3).

2. Find where the line r = (1, 2, −4) + λ(2, −1, 2) meets the xy-plane.

4 The angle between two lines

Use the direction vectors.

The angle between two lines is the angle between their direction vectors, found with the scalar product. If the result is obtuse, the acute angle between the lines is 180° minus it; using the absolute value |b1·b2| gives the acute angle directly.

WORKED EXAMPLE 4 Angle between lines (standard)

Find the acute angle between r = (1, 0, 2) + λ(2, −1, 2) and r = (0, 3, 1) + μ(1, 1, −1).

Solution

b1·b2 = 2 − 1 − 2 = −1; |b1| = 3, |b2| = √3A1

cos θ = 1/(3√3)M1

θ = 78.9°A1

Common mistake

Using the position vectors (a1, a2) instead of the direction vectors. The position vectors only locate the lines.

Practice exercise 4

1. Find the acute angle between lines with directions (1, 2, 2) and (2, −2, 1).

2. Show that the lines with directions (3, −1, 2) and (1, 1, −1) are perpendicular.

5 Intersecting, parallel and skew lines

Three possibilities in 3D.

In 2D, two lines are either parallel or intersect. In 3D there is a third possibility: skew lines are not parallel and do not meet.

Method: (1) If the direction vectors are multiples, the lines are parallel (check whether they coincide by testing a point). (2) Otherwise, set the two parametric forms equal, using different parameters λ and μ. Solve two of the three equations for λ and μ, then substitute into the third equation. If it is satisfied, the lines intersect; find the point. If not, the lines are skew.

Directions parallel?	Equations consistent?	Relationship
yes	common point	coincident (same line)
yes	no common point	parallel
no	yes	intersecting
no	no	skew

WORKED EXAMPLE 5 Testing for intersection (standard)

Determine whether L1: r = (1, 2, 3) + λ(1, 1, 2) and L2: r = (3, 1, 2) + μ(1, 2, 3) intersect.

Solution

x: 1 + λ = 3 + μ; y: 2 + λ = 1 + 2μM1

Subtract: 1 = −2 + μ, μ = 3, λ = 5A1

z: 3 + 10 = 13 and 2 + 9 = 11. Not equal, so the lines do not intersect: they are skewR1

This example shows why the third equation must always be checked.

WORKED EXAMPLE 6 Finding an intersection (exam level)

Find the intersection of L1: r = (1, 0, −1) + λ(2, 1, 3) and L2: r = (4, 4, 5) + μ(1, −2, 0).

Solution

1 + 2λ = 4 + μ and λ = 4 − 2μM1

1 + 8 − 4μ = 4 + μ: μ = 1, λ = 2A1

z: −1 + 6 = 5 ✓; point (5, 2, 5)R1 A1

Practice exercise 5

1. Determine whether r = (0, 1, 2) + λ(1, −1, 1) and r = (2, 0, 3) + μ(−2, 2, −2) are parallel or coincident.

2. Show that r = (1, 1, 0) + λ(1, 0, 1) and r = (0, 2, 1) + μ(0, 1, 1) are skew.

3. Find k if r = (1, 2, k) + λ(1, 1, 1) and r = (3, 1, 0) + μ(0, 1, 2) intersect.

6 Shortest distance from a point to a line

The foot of the perpendicular.

The shortest distance from a point P to a line is along the perpendicular. Let F be a general point on the line, F = a + λb. Then PF is perpendicular to the line when PF·b = 0. Solve for λ to find the foot of the perpendicular F, then the distance |PF|.

WORKED EXAMPLE 7 Foot of the perpendicular (exam level)

Find the shortest distance from P(3, 4, 1) to the line r = (1, 0, 2) + λ(1, 2, −1).

Solution

F = (1 + λ, 2λ, 2 − λ); PF = (λ − 2, 2λ − 4, 1 − λ)M1

PF·(1, 2, −1) = λ − 2 + 4λ − 8 − 1 + λ = 6λ − 11 = 0, λ = 11/6A1

PF = (−1/6, −1/3, −5/6), |PF| = √(30)/6 = 0.913A1

Practice exercise 6

1. Find the foot of the perpendicular from (0, 0, 0) to r = (2, 1, 0) + λ(1, −1, 1).

2. Find the distance from (1, 2, 3) to the x-axis.

7 Lines in kinematics

Position = start + time × velocity.

An object moving with constant velocity v from initial position a has position r = a + tv at time t. The speed is |v|. Two objects collide only if they are at the same point at the same time t; if their paths cross at different times, they do not collide.

WORKED EXAMPLE 8 Ship position (standard)

A ship starts at (2, −1) km and moves with velocity (3, 4) km/h. Find its speed and its position after 2.5 hours.

Solution

Speed |(3, 4)| = 5 km/hA1

r = (2, −1) + 2.5(3, 4) = (9.5, 9) kmA1

WORKED EXAMPLE 9 Collision check (exam level)

Ship A: r = (0, 0) + t(4, 3); ship B: r = (10, −5) + t(−1, 5). Do they collide?

Solution

Equate: 4t = 10 − t gives t = 2M1

y: 3(2) = 6; −5 + 5(2) = 5A1

6 ≠ 5, so they are never at the same point at the same time: no collisionR1

Practice exercise 7

1. A drone starts at (0, 0, 50) m and moves with velocity (4, 3, −1) m/s. When does it reach the ground (z = 0), and where?

2. Determine whether particles r = (1, 2) + t(2, 1) and r = (7, 1) + t(−1, 2) collide.

8 Closest approach

Minimize the distance between two moving objects.

The displacement between two moving objects is d(t) = rB(t) − rA(t). The distance |d(t)| is smallest when d(t) is perpendicular to the relative velocity, or equivalently when |d(t)|2 (a quadratic in t) is minimized. Solve for t, then find the minimum distance.

WORKED EXAMPLE 10 Closest approach (exam level)

Ship A: r = (0, 0) + t(4, 3); ship B: r = (10, −5) + t(−1, 5), in km and hours. Find the minimum distance between the ships.

Solution

d(t) = (10 − 5t, −5 + 2t)M1

|d|2 = (10 − 5t)2 + (−5 + 2t)2 = 29t2 − 120t + 125A1

Minimum at t = 120/58 = 2.07 h: |d| = √(125 − 3600/29) = 0.928 kmM1 A1

Practice exercise 8

1. Particles move with rA = (0, 0) + t(1, 2) and rB = (5, 0) + t(0, 1). Find their minimum separation.

9 Lines in two dimensions with vectors

Linking to y = mx + c.

A 2D line r = (a1, a2) + λ(b1, b2) has gradient b2/b1. Eliminating λ from x = a1 + λb1, y = a2 + λb2 gives the Cartesian equation.

WORKED EXAMPLE 11 Converting (standard)

Find the Cartesian equation of r = (1, 3) + λ(2, −5).

Solution

x = 1 + 2λ, y = 3 − 5λM1

λ = (x − 1)/2: y = 3 − 5(x − 1)/2: 5x + 2y = 11A1

Practice exercise 9

1. Write y = 3x − 2 in vector form.

2. Find the intersection of r = (0, 1) + λ(1, 2) and r = (4, 0) + μ(−1, 1).

10 Choosing direction and position vectors

Many equations, one line.

Any point on the line can serve as the position vector and any non-zero multiple of the direction can serve as the direction vector. To check whether two equations describe the same line, check that the directions are parallel and that a point of one lies on the other.

WORKED EXAMPLE 12 Same line? (standard)

Do r = (1, 2, 3) + λ(2, 0, −2) and r = (5, 2, −1) + μ(−1, 0, 1) describe the same line?

Solution

Directions parallelA1

(5, 2, −1) = (1, 2, 3) + 2(2, 0, −2) ✓M1

Yes, same lineR1

Practice exercise 10

1. Give two different vector equations for the line through (0, 1, 2) and (2, 5, 0).

11 Intersection problems with unknowns

Find the constant that makes lines meet.

If a line contains an unknown, set up the parametric equations equal, solve two of them for the parameters, and substitute into the third to find the unknown that makes them consistent.

WORKED EXAMPLE 13 Unknown constant (exam level)

L1: r = (1, 0, 2) + λ(1, 1, 0); L2: r = (3, 1, k) + μ(0, 1, 1). Find k so that the lines intersect, and the point.

Solution

x: 1 + λ = 3: λ = 2A1

y: 2 = 1 + μ: μ = 1A1

z: 2 = k + 1: k = 1; point (3, 2, 2)A1 A1

Practice exercise 11

1. Show that there is no value of a for which r = (2, a, 1) + λ(1, 0, 1) meets r = (0, 1, 0) + μ(1, 1, 1).

12 Angle between lines: further practice

Acute angles only.

The angle between two lines uses only their direction vectors and is taken as acute (use |b1·b2|). The lines need not intersect: the angle between skew lines is defined in the same way.

WORKED EXAMPLE 14 Skew lines angle (standard)

Find the acute angle between r = λ(1, 2, 2) and r = (1, 0, 0) + μ(2, −1, 2).

Solution

Dot product 2 − 2 + 4 = 4; magnitudes 3 and 3A1

cos θ = 4/9: θ = 63.6°A1

Practice exercise 12

1. Find the angle between the line r = λ(1, 1, 0) and the z-axis.

13 Foot of the perpendicular and reflections

Using the scalar product.

The foot F of the perpendicular from P to a line satisfies PF·b = 0. The reflection P′ of P in the line is P′ = 2F − P (F is the midpoint of PP′).

WORKED EXAMPLE 15 Reflection (exam level)

Find the reflection of P(4, 1, 0) in the line r = (0, 0, 0) + λ(1, 1, 1).

Solution

F = (λ, λ, λ); PF·(1, 1, 1) = 3λ − 5 = 0M1 A1

F = (5/3, 5/3, 5/3)A1

P′ = 2F − P = (−2/3, 7/3, 10/3)A1

Practice exercise 13

1. Find the distance from (2, 3, 4) to the line r = λ(1, 0, 0).

14 Shortest distance between skew lines

Using a common perpendicular (extension).

The shortest distance between two skew lines is along a line perpendicular to both. One method: take general points P(λ) and Q(μ) on each line and require PQ to be perpendicular to both direction vectors; solve for λ and μ, then find |PQ|.

WORKED EXAMPLE 16 Skew distance (exam level)

Find the shortest distance between r = (1, 0, 0) + λ(0, 1, 0) and r = (0, 0, 2) + μ(1, 0, 0).

Solution

PQ = (μ − 1, −λ, 2)M1

PQ·(0, 1, 0) = 0: λ = 0; PQ·(1, 0, 0) = 0: μ = 1A1

|PQ| = |(0, 0, 2)| = 2A1

Practice exercise 14

1. Explain why the shortest distance between two intersecting lines is 0.

15 Kinematics: collision and closest approach

Same place, same time.

Two objects collide only if they are at the same point at the same time. The closest approach occurs when the relative position vector is perpendicular to the relative velocity, or when the squared distance (a quadratic in t) is minimized.

WORKED EXAMPLE 17 Closest approach (exam level)

A: r = (0, 0) + t(3, 4); B: r = (10, 2) + t(−1, 3). Find the minimum distance and when it occurs.

Solution

d(t) = (10 − 4t, 2 − t)M1

|d|2 = 17t2 − 84t + 104: minimum at t = 42/17 = 2.47A1

|d| = √(104 − 422/17) = 0.485A1

Practice exercise 15

1. Do particles r = (1, 1) + t(2, 0) and r = (5, −3) + t(0, 2) collide?

16 Lines and geometric shapes

Triangles and parallelograms in 3D.

Lines define edges of shapes. Use vector equations to find vertices (as intersections), to test whether points are collinear, and to find angles and lengths.

WORKED EXAMPLE 18 Triangle from lines (standard)

Lines L1, L2 meet at A(1, 2, 3). B = (3, 3, 5) lies on L1 and C = (1, 5, 3) on L2. Find angle BAC.

Solution

AB = (2, 1, 2), AC = (0, 3, 0)A1

cos A = 3/(3 × 3) = 1/3: 70.5°M1 A1

Practice exercise 16

1. Find the area of triangle ABC above.

17 Mixed review: worked problems

Bringing the ideas together.

WORKED EXAMPLE 19 Mixed problem (exam level)

L1: r = (2, 1, 0) + λ(1, 2, 1); L2: r = (5, 4, 1) + μ(−1, 1, 2). (a) Determine whether the lines intersect. (b) Find the angle between them.

Solution

(a) 2 + λ = 5 − μ and 1 + 2λ = 4 + μ: λ = 2, μ = 1M1 A1

z: 0 + 2 = 2 but 1 + 2(1) = 3: not equal, so the lines are skewR1

(b) dot product −1 + 2 + 2 = 3; magnitudes √6, √6: cos θ = 1/2: 60°A1

Practice exercise 17

1. Find the point on r = (1, 2, 3) + λ(1, 0, −1) closest to the origin.

18 Additional exam-style practice

Paper 2.

WORKED EXAMPLE 20 Paper 2 question (exam level)

A plane leaves A(0, 0, 0) with velocity (200, 150, 5) km/h. A drone starts at (600, 300, 20) with velocity (−100, 0, 0). (a) Find the plane's speed. (b) Do they collide? (c) Find the minimum distance using technology.

Solution

(a) √(40 000 + 22 500 + 25) = 250 km/hA1

(b) x: 200t = 600 − 100t, t = 2; y: 300 = 300 ✓; z: 10 vs 20: no collisionM1 R1

Practice exercise 18

1. Find the angle the plane's path makes with the horizontal.

19 Vector equations with technology

Solving parameters with a GDC.

To test for intersection, use the GDC simultaneous solver for two of the parameter equations, then check the third. For closest approach, graph |d(t)|2 and find its minimum.

WORKED EXAMPLE 21 Check (standard)

Use technology to confirm the closest approach time in the kinematics example.

Solution

Minimum of 17t2 − 84t + 104 at t = 2.47A1

GDC tip

Enter |d(t)|2 as a function of t and use the minimum tool. The minimum distance is the square root of the minimum value.

20 Interpreting parameters

What does λ tell us?

In r = a + λb, λ = 0 gives the starting point, λ = 1 the point a + b, and negative λ points in the opposite direction. If |b| = 1, |λ| is the distance from a. In kinematics, the parameter is time.

WORKED EXAMPLE 22 Distance along a line (standard)

On r = (1, 1, 1) + λ(2, 1, 2), find the points 6 units from (1, 1, 1).

Solution

|λb| = 3|λ| = 6: λ = ±2M1

(5, 3, 5) and (−3, −1, −3)A1

Practice exercise 19

1. Find the point on r = (0, 2, −1) + λ(3, 4, 0) at distance 10 from (0, 2, −1), with λ > 0.

21 Exam-style questions with markscheme

Attempt each question under timed conditions before reading the markscheme.

Question 1 (Paper 1 style) (about 15 minutes)

L1: r = (2, 1, −1) + λ(1, −1, 2) and L2: r = (1, 4, 1) + μ(2, −3, 1).

(a) Determine whether L1 and L2 intersect. [5]

(b) Find the acute angle between the lines. [3]

Markscheme

Part	Answer	Marks
(a)	2 + λ = 1 + 2μ and 1 − λ = 4 − 3μ	M1
	add: 3 = 5 − μ, μ = 2, λ = 3	A1 A1
	z: −1 + 6 = 5 and 1 + 2 = 3: not equal	R1
	the lines do not intersect (skew)	A1
(b)	b1·b2 = 2 + 3 + 2 = 7	A1
	cos θ = 7/(√6 √14)	M1
	40.2°	A1

Examiner's comments. Candidates who did not check the third equation wrongly concluded that the lines intersect.

Question 2 (Paper 2 style) (about 12 minutes)

A boat leaves harbour O at noon with velocity (6, 8) km/h. A lighthouse is at L(20, 10) km.

(a) Find the speed of the boat. [1]

(b) Write the position of the boat t hours after noon. [1]

Markscheme

Part	Answer	Marks
(a)	10 km/h	A1
(b)	r = t(6, 8)	A1
(c)	d = (6t − 20, 8t − 10)	M1
	d·(6, 8) = 0: 100t − 200 = 0	M1
	t = 2: 2 p.m.	A1
	d = (−8, 6)	A1
	10 km	A1

Examiner's comments. Setting the displacement perpendicular to the velocity gives the closest point directly.

22 Common misconceptions and links

Students often think…	Actually…
two non-parallel lines in 3D always meet	they may be skew
paths crossing means a collision	the times must also match
use the same parameter for both lines	use λ and μ
the angle uses position vectors	use direction vectors

How 3.14 connects to the rest of the course

Topic	Connection
3.12 Vectors	scalar product
3.16 Planes	lines meeting planes
1.16 Systems	solving for parameters
5.8 Optimization	minimizing distance

23 Check your understanding

Section A: multiple choice

1. A direction vector of r = (1, 2, 3) + λ(4, 5, 6) is

A. (4, 5, 6) B. (1, 2, 3) C. (5, 7, 9) D. (3, 3, 3)

2. Skew lines are

A. parallel B. perpendicular C. not parallel and do not intersect D. coincident

3. Two objects collide if

A. their paths cross B. they are at the same point at the same time C. their speeds are equal D. their velocities are parallel

4. Speed is

A. the velocity B. the direction C. the position D. the magnitude of velocity

Section B: structured

5. Find the Cartesian equation of the line through (2, −1, 0) and (5, 1, 4). [3]

6. Find the distance from (2, 3, 1) to r = λ(1, 1, 1). [4]

Answers

1. A 2. C 3. B 4. D

5. (x − 2)/3 = (y + 1)/2 = z/4.

6. λ = 2: foot (2, 2, 2): distance √(0 + 1 + 1) = √2.

24 Summary and topic map

Idea	Remember
Vector form	r = a + λb
Parametric	x = a1 + λb1, ...
Cartesian	(x − a1)/b1 = (y − a2)/b2 = (z − a3)/b3
Angle	\|b1·b2\|/(\|b1\|\|b2\|)
Relationships	intersecting, parallel, coincident, skew
Point to line	PF·b = 0
Kinematics	r = a + tv; collisions need equal times

Before the exam, make sure you…

•use different parameters
•check the third equation
•use direction vectors for angles
•minimize distance with a dot product or quadratic

Vector Equations of Lines on one page

Cover this page with a sheet of paper and try to rebuild it from memory. Whatever you can't recall is what to revise next.

Figure 2. Topic map for Vector Equations of Lines.

25 Answers to practice exercises

Check your working against these answers. Full methods are expected in the exam.

Practice exercise 1

1. r = (2, 0, 5) + λ(1, 0, −3)

2. r = (1, 1, 1) + λ(2, −2, 3)

3. r = λ(0, 0, 1)

Practice exercise 2

1. (x − 1)/3 = (y − 4)/(−1) = (z + 2)/2

2. x = ½ + 2λ, y = −3 + 2λ, z = 3 − λ: r = (½, −3, 3) + λ(2, 2, −1)

Practice exercise 3

1. λ = 2: a = 4, b = −6

2. z = 0 when λ = 2: (5, 0, 0)

Practice exercise 4

1. dot product 0: 90°

2. 3 − 1 − 2 = 0

Practice exercise 5

1. directions parallel; (2, 0, 3) is not on the first line (λ = 2 gives (2, −1, 4)): parallel, distinct

2. x: 1 + λ = 0, λ = −1; y: 1 = 2 + μ, μ = −1; z: −1 ≠ 0: skew

3. λ = 2, μ = 3: k + 2 = 6, k = 4

Practice exercise 6

1. λ = −1/3: (5/3, 4/3, −1/3)

2. √13

Practice exercise 7

1. t = 50 s at (200, 150, 0)

2. x: 1 + 2t = 7 − t gives t = 2; y: 2 + 2 = 4 but 1 + 4 = 5, so they do not collide

Practice exercise 8

1. d = (5 − t, −t): |d|2 = 2t2 − 10t + 25, min at t = 2.5: √12.5 = 3.54

Practice exercise 9

1. r = (0, −2) + λ(1, 3)

2. λ = 1, μ = 3: (1, 3)

Practice exercise 10

1. r = (0, 1, 2) + λ(2, 4, −2); r = (2, 5, 0) + μ(1, 2, −1)

Practice exercise 11

1. x gives μ = 2 + λ and z gives μ = 1 + λ: a contradiction for every a

Practice exercise 12

1. 90°

Practice exercise 13

1. 5

Practice exercise 14

1. they share a point

Practice exercise 15

1. at t = 2 both are at (5, 1): yes

Practice exercise 16

1. ½|AB × AC| = ½|(−6, 0, 6)| = 4.24

Practice exercise 17

1. λ = 1: (2, 2, 2)

Practice exercise 18

1. tan θ = 5/250: 1.15°

Practice exercise 19

1. λ = 2: (6, 10, −1)

DP MATHEMATICS AA · TOPIC 3: GEOMETRY AND TRIGONOMETRY

3.16

The Vector Product and Planes

Full notes for HL · Syllabus AHL 3.16, 3.17, 3.18

Contents

1	The vector (cross) product
2	Properties and geometric meaning
3	Areas of parallelograms and triangles
4	Equations of a plane
5	Finding the equation of a plane
6	Intersection of a line and a plane
7	Angles involving planes
8	Intersection of two and three planes
9	Distances from a point to a plane
10	Computing cross products reliably
11	Areas and volumes with cross products
12	Converting between forms of a plane
13	Planes through given points and lines
14	Line and plane: intersection, parallel or contained
15	Angles between lines and planes, and between planes
16	Distance from a point to a plane
17	Intersection of two planes
18	Three planes
19	Reflections in planes
20	Mixed review: worked problems
21	Additional exam-style practice
22	Exam-style questions with markscheme
23	Common misconceptions and links
24	Check your understanding
25	Summary and topic map
26	Answers to practice exercises

These are full teaching notes. Read each section in order, try the Think about it question before you read on, and work through every worked example before attempting the Quick check.

Before you start

Rate yourself on each statement now, and again after you've worked through the notes. Shade R (not yet), A (getting there) or G (confident).

I can…	R	A	G
calculate the vector product v × w and know that it is perpendicular to both v and w
use \|v × w\| = \|v\|\|w\| sin θ to find areas of parallelograms and triangles
write the equation of a plane in vector, parametric, normal (r·n = a·n) and Cartesian (ax + by + cz = d) forms
find the equation of a plane through three points or containing a line and a point
find the intersection of a line and a plane
find the angle between a line and a plane and between two planes
find the line of intersection of two planes and interpret systems of three planes
find the distance from a point to a plane

This booklet covers Additional Higher Level (AHL) content (AHL 3.16, 3.17, 3.18). It is studied by HL students only.

Guiding questions

How can a single vector describe the orientation of a flat surface in space? How do lines and planes meet?

How this topic is examined

Vector product and planes form the final part of the HL vectors topic and appear as long questions on Papers 1 and 2: finding a plane through three points, intersecting a line with a plane, angles, and the intersection of planes (linked to systems of equations, 1.16).

Check a normal vector by confirming its dot products with the plane's direction vectors are zero.

Command term	What the examiner wants
Write down	an answer with little or no working; it should be obtainable by inspection
Find / Calculate / Determine	obtain the answer showing relevant working
Solve	find the values of the unknowns, using algebra or technology as allowed
Show that	obtain the given result; every step must be shown and the result must not be assumed
Hence / Hence or otherwise	use the previous result (or another valid method)
Sketch	a diagram showing the general shape and key features, labelled
Prove	a formal, rigorous argument from accepted facts
Justify / Explain	give valid reasons or evidence

How to use these notes

Each section teaches one part of the syllabus. You'll see these boxes throughout; each has one job:

Box	What it's for
Think about it	a question to get you thinking before a section
Why it works	a derivation or proof of a key result, so you understand where formulas come from
Key point	a result or idea you must not forget
Common mistake	an error examiners see every year, and how to avoid it
GDC tip	how to use your graphic display calculator efficiently on Paper 2
In the real world	where the mathematics is used outside the classroom
Worked example	a full solution with IB-style marks (M1 method, A1 answer, R1 reasoning, AG answer given); examples are graded basic, standard and exam level
Practice exercise	graded questions for each section; full answers are at the end of the booklet
Quick check	short questions to check the key ideas

1 The vector (cross) product

A vector perpendicular to two others.

Think about it

Two different vectors lying in a flat table top define the table's plane. How could you find a vector sticking straight up out of the table?

The vector product (cross product) of v and w is a vector:

WORKED EXAMPLE 1 Computing a vector product (basic)

Find (2, 1, −1) × (1, 3, 2).

Solution

i: 1(2) − (−1)(3) = 5; j: (−1)(1) − 2(2) = −5; k: 2(3) − 1(1) = 5M1 A1

= (5, −5, 5); check (5, −5, 5)·(2, 1, −1) = 0 ✓R1

GDC tip

Most GDCs have a cross-product function in the vector/matrix menu. On Paper 1 compute by hand and check your answer with two dot products.

Practice exercise 1

1. Find i × j and j × i.

2. Find (1, 0, 2) × (3, −1, 1).

3. Find a vector perpendicular to both (1, 1, 0) and (0, 2, 1).

2 Properties and geometric meaning

Perpendicular, with length |v||w| sin θ.

Direction: v × w is perpendicular to both v and w, following the right-hand rule (fingers from v towards w, thumb gives v × w). Magnitude: |v × w| = |v||w| sin θ. Properties: w × v = −(v × w) (not commutative); v × v = 0; v and w are parallel if and only if v × w = 0; v × (w + u) = v × w + v × u.

	Scalar product v·w	Vector product v × w
Result	a number	a vector
Size	\|v\|\|w\| cos θ	\|v\|\|w\| sin θ
Zero when	perpendicular	parallel
Order	commutative	anticommutative

3 Areas of parallelograms and triangles

Magnitude of the cross product.

WORKED EXAMPLE 2 Area of a triangle (standard)

Find the area of the triangle with vertices A(1, 0, 1), B(3, 1, 0) and C(2, 3, 2).

Solution

AB = (2, 1, −1), AC = (1, 3, 1)A1

AB × AC = (1 + 3, −1 − 2, 6 − 1) = (4, −3, 5)M1 A1

Area = ½√(16 + 9 + 25) = ½√50 = 3.54A1

Practice exercise 2

1. Find the area of the parallelogram spanned by (1, 2, 0) and (3, 0, 4).

2. Find the area of the triangle P(0, 0, 0), Q(1, 2, 3), R(3, 2, 1).

4 Equations of a plane

A point and a normal, or a point and two directions.

A plane is determined by a point A on it and a normal vector n perpendicular to it. Any point R in the plane satisfies AR·n = 0, so (r − a)·n = 0:

Figure 1. The normal n is perpendicular to every line AR in the plane.

Alternatively, a plane through A containing two non-parallel direction vectors b and c has vector (parametric) equation r = a + λb + μc. The normal is then n = b × c. Note that in the Cartesian form ax + by + cz = d, the coefficients of x, y and z form a normal vector.

Practice exercise 3

1. Write down a normal vector of 2x − y + 3z = 7.

2. Find the Cartesian equation of the plane through (1, 2, −1) with normal (3, 0, 4).

3. Does (2, 1, 1) lie on x + 2y − z = 3?

5 Finding the equation of a plane

Use the cross product for the normal.

WORKED EXAMPLE 3 Plane through three points (standard)

Find the Cartesian equation of the plane through A(1, 0, 1), B(3, 1, 0) and C(2, 3, 2).

Solution

AB × AC = (4, −3, 5) (from the previous section) is a normalM1 A1

4x − 3y + 5z = 4(1) − 0 + 5(1) = 9M1

4x − 3y + 5z = 9A1

WORKED EXAMPLE 4 Plane containing a line and a point (exam level)

Find the plane containing the line r = (1, 2, 0) + λ(1, −1, 2) and the point P(3, 0, 1).

Solution

Directions: (1, −1, 2) and AP = (2, −2, 1)M1

n = (1, −1, 2) × (2, −2, 1) = (−1 + 4, 4 − 1, −2 + 2) = (3, 3, 0)A1

Use n = (1, 1, 0): x + y = 3A1

Practice exercise 4

1. Find the plane through (0, 0, 0), (1, 1, 0) and (0, 1, 1).

2. Find the plane through (2, 1, 3) parallel to 2x − y + z = 5.

3. Find the plane containing the parallel lines r = (1, 0, 0) + λ(1, 1, 1) and r = (0, 1, 0) + μ(1, 1, 1).

6 Intersection of a line and a plane

Substitute the parametric line.

Substitute the parametric coordinates of the line into the plane's Cartesian equation and solve for λ. A unique solution gives the intersection point. If the equation becomes impossible (such as 0 = 5), the line is parallel to the plane and does not meet it; if it becomes 0 = 0 for all λ, the line lies in the plane. The line is parallel to the plane exactly when its direction is perpendicular to the normal: b·n = 0.

WORKED EXAMPLE 5 Line meets plane (standard)

Find where r = (1, 2, 3) + λ(2, −1, 1) meets 3x + y − z = 10.

Solution

3(1 + 2λ) + (2 − λ) − (3 + λ) = 10M1

2 + 4λ = 10, λ = 2A1

Point (5, 0, 5)A1

Practice exercise 5

1. Find the intersection of r = (0, 1, 2) + λ(1, 1, 1) with x + 2y + 3z = 20.

2. Show that r = (1, 1, 1) + λ(1, −1, 0) is parallel to x + y + z = 5 and does not lie in it.

7 Angles involving planes

Normals and directions.

Between two planes: the angle between their normals (take the acute angle). Between a line and a plane: if θ is the angle between the line's direction b and the normal n, the angle between the line and the plane is 90° − θ; equivalently, sin φ = |b·n|/(|b||n|).

WORKED EXAMPLE 6 Line and plane (exam level)

Find the angle between r = (1, 0, 2) + λ(1, 2, 2) and x + y − z = 4.

Solution

b·n = 1 + 2 − 2 = 1; |b| = 3, |n| = √3A1

sin φ = 1/(3√3)M1

φ = 11.1°A1

Common mistake

Giving the angle between the line and the normal as the answer. Subtract it from 90° (or use sine).

Practice exercise 6

1. Find the acute angle between the planes x + y + z = 1 and 2x − y + 3z = 4.

2. Find the angle between the line with direction (0, 0, 1) and the plane x + y + z = 0.

8 Intersection of two and three planes

Lines and points.

Two planes that are not parallel meet in a line. Its direction is n1 × n2 (perpendicular to both normals). To find a point on it, set one coordinate (say z = 0) and solve the two plane equations; or solve the system with a parameter (1.16).

Three planes may meet at a single point (unique solution), in a line (sheaf), not at all (parallel planes or a triangular prism), or coincide. These correspond to the solutions of a 3 × 3 system (1.16).

WORKED EXAMPLE 7 Line of intersection (exam level)

Find the line of intersection of x + y + z = 6 and 2x − y + z = 3.

Solution

Direction (1, 1, 1) × (2, −1, 1) = (2, 1, −3)M1 A1

Set z = 0: x + y = 6, 2x − y = 3, so x = 3, y = 3M1

r = (3, 3, 0) + λ(2, 1, −3)A1

Practice exercise 7

1. Find the line of intersection of x − y = 1 and y + z = 2.

2. Describe the arrangement of x + y + z = 1, x + y + z = 2 and x − y = 0.

9 Distances from a point to a plane

Drop a perpendicular.

To find the shortest distance from a point P to a plane, take the line through P in the direction of the normal: r = p + λn. Find where it meets the plane (the foot F), then calculate |PF|. Equivalently, the distance is |ax0 + by0 + cz0 − d|/√(a2 + b2 + c2).

WORKED EXAMPLE 8 Distance to a plane (exam level)

Find the shortest distance from P(4, 1, 3) to the plane 2x − y + 2z = 3.

Solution

Line r = (4, 1, 3) + λ(2, −1, 2); substitute: 2(4 + 2λ) − (1 − λ) + 2(3 + 2λ) = 3M1

13 + 9λ = 3, λ = −10/9A1

Distance = |λ||n| = (10/9)(3) = 10/3A1

Practice exercise 8

1. Find the distance from the origin to x + 2y + 2z = 9.

2. Find the foot of the perpendicular from (1, 1, 1) to x + y + z = 6.

10 Computing cross products reliably

The determinant pattern.

v × w = (v2w3 − v3w2, v3w1 − v1w3, v1w2 − v2w1). Always check the result by confirming that its dot product with both v and w is zero.

WORKED EXAMPLE 9 Cross product (basic)

Find (3, −1, 2) × (1, 4, −2) and check.

Solution

(2 − 8, 2 + 6, 12 + 1) = (−6, 8, 13)M1 A1

Check: −18 − 8 + 26 = 0; −6 + 32 − 26 = 0 ✓R1

Practice exercise 9

1. Find (1, 2, 3) × (4, 5, 6).

2. Find (2, 0, −1) × (0, 3, 1).

11 Areas and volumes with cross products

Parallelograms, triangles, parallelepipeds.

Area of parallelogram = |a × b|; triangle = ½|a × b|. The volume of a parallelepiped with edges a, b, c is |a·(b × c)| (the scalar triple product); a tetrahedron has one sixth of this volume.

WORKED EXAMPLE 10 Tetrahedron (exam level)

Find the volume of the tetrahedron with vertices O, A(1, 0, 0), B(0, 2, 0), C(1, 1, 3).

Solution

b × c = (6, 0, −2)A1

a·(b × c) = 6M1

Volume = 6/6 = 1A1

Practice exercise 10

1. Find the area of the triangle with vertices (1, 1, 1), (2, 3, 1), (1, 2, 4).

2. Show that (1, 2, 3), (2, 1, 0), (3, 3, 3) are coplanar using the triple product.

12 Converting between forms of a plane

Vector, normal and Cartesian.

From r = a + λb + μc: normal n = b × c, then r·n = a·n gives the Cartesian form. From the Cartesian form ax + by + cz = d, find three points (for example, intercepts) to write a vector form.

WORKED EXAMPLE 11 Vector to Cartesian (standard)

Find the Cartesian equation of r = (1, 0, 2) + λ(1, 1, 0) + μ(0, 1, 1).

Solution

n = (1, −1, 1)A1

x − y + z = 1 + 0 + 2M1

x − y + z = 3A1

Practice exercise 11

1. Write 2x + y − z = 4 in vector form.

2. Find the intercepts of 3x − 2y + 6z = 12.

13 Planes through given points and lines

Two directions and a point.

A plane needs a point and a normal. From three points, take two direction vectors in the plane and cross them. From a line and a point not on it, use the line's direction and the vector from a point on the line to the given point.

WORKED EXAMPLE 12 Perpendicular plane (standard)

Find the plane through (1, 2, −1) perpendicular to the line r = λ(2, −1, 3).

Solution

n = (2, −1, 3)A1

2x − y + 3z = −3M1 A1

Practice exercise 12

1. Find the plane containing the points (1, 0, 0), (0, 1, 0), (0, 0, 1).

2. Find the plane containing the two intersecting lines r = λ(1, 1, 0) and r = μ(0, 1, 1).

14 Line and plane: intersection, parallel or contained

Three cases.

Substitute the line into the plane. A unique λ gives a single point. If the equation becomes a false statement, the line is parallel to the plane; if it becomes 0 = 0, the line lies in the plane. Parallel lines satisfy b·n = 0.

WORKED EXAMPLE 13 Line in the plane (standard)

Show that r = (1, 1, 1) + λ(1, −1, 0) lies in x + y + z = 3.

Solution

(1 + λ) + (1 − λ) + 1 = 3 for all λM1

Line lies in the planeR1

Practice exercise 13

1. Find the intersection of r = (2, 0, 1) + λ(1, 2, −1) with 2x − y + z = 9.

2. Determine whether r = (0, 0, 1) + λ(1, 1, 2) meets x + y − z = 5.

15 Angles between lines and planes, and between planes

Sine for lines, cosine for planes.

Line-plane: sin φ = |b·n|/(|b||n|). Plane-plane: cos θ = |n1·n2|/(|n1||n2|). Always give the acute angle.

WORKED EXAMPLE 14 Plane-plane (standard)

Find the angle between x + 2y + 2z = 5 and 2x − y + 2z = 1.

Solution

n1·n2 = 2 − 2 + 4 = 4; |n1| = |n2| = 3A1

cos θ = 4/9: θ = 63.6°A1

Practice exercise 14

1. Find the angle between the line r = λ(1, 0, 1) and the plane z = 0.

2. Find the angle between the planes x = 0 and x + y = 0.

16 Distance from a point to a plane

Along the normal.

The perpendicular distance from (x0, y0, z0) to ax + by + cz = d is |ax0 + by0 + cz0 − d|/√(a2 + b2 + c2). The foot of the perpendicular is found by intersecting the normal line through the point with the plane.

WORKED EXAMPLE 15 Distance and foot (exam level)

Find the distance from P(3, 1, 2) to 2x − 2y + z = 1 and the foot of the perpendicular.

Solution

|6 − 2 + 2 − 1|/3 = 5/3M1 A1

Line (3, 1, 2) + t(2, −2, 1): 2(3 + 2t) − 2(1 − 2t) + 2 + t = 1: t = −5/9M1

Foot (17/9, 19/9, 13/9)A1

Practice exercise 15

1. Find the distance between the parallel planes x + 2y + 2z = 3 and x + 2y + 2z = 12.

17 Intersection of two planes

A line.

The line of intersection has direction n1 × n2. Find a point on it by setting one variable (often z = 0) and solving the two plane equations.

WORKED EXAMPLE 16 Line of intersection (standard)

Find the line of intersection of x + y − z = 2 and 2x − y + z = 1.

Solution

Direction (1, 1, −1) × (2, −1, 1) = (0, −3, −3) ∝ (0, 1, 1)M1 A1

z = 0: x + y = 2, 2x − y = 1: x = 1, y = 1A1

r = (1, 1, 0) + λ(0, 1, 1)A1

Practice exercise 16

1. Find the line of intersection of z = 0 and x + y + z = 4.

18 Three planes

Point, line or none.

Solve the three Cartesian equations simultaneously (1.16). A unique solution is a point. If the system is dependent, the planes share a line (or are coincident). If inconsistent, there is no common point: check whether two planes are parallel (normals proportional) or the planes form a triangular prism.

WORKED EXAMPLE 17 Classification (exam level)

Describe the intersection of x + y + z = 1, x − y + z = 3, 2x + 2z = 4.

Solution

R1 + R2 = 2x + 2z = 4 = R3: dependentM1

Planes meet in a line: x + z = 2, y = −1A1

r = (2, −1, 0) + λ(−1, 0, 1)A1

Practice exercise 17

1. Find the point of intersection of x + y + z = 6, x − y = 0, z = 2.

19 Reflections in planes

Twice the distance along the normal.

To reflect P in a plane, find the foot F of the perpendicular from P; then P′ = 2F − P. For a line, reflect two points (or reflect one point and the direction vector).

WORKED EXAMPLE 18 Reflection (exam level)

Reflect P(1, 2, 3) in the plane x + y + z = 3.

Solution

Line P + t(1, 1, 1): 6 + 3t = 3, t = −1; F = (0, 1, 2)M1 A1

P′ = 2F − P = (−1, 0, 1)A1

Practice exercise 18

1. Reflect the origin in the plane 2x + y − 2z = 9.

20 Mixed review: worked problems

Bringing the ideas together.

WORKED EXAMPLE 19 Mixed problem (exam level)

Π: x − 2y + 2z = 4 and L: r = (1, 1, 1) + λ(2, 1, −1). (a) Find the intersection of L and Π. (b) Find the acute angle between L and Π.

Solution

(a) (1 + 2λ) − 2(1 + λ) + 2(1 − λ) = 4: 1 − 2λ = 4, λ = −3/2: (−2, −1/2, 5/2)M1 A1

(b) b·n = 2 − 2 − 2 = −2; |b| = √6, |n| = 3: sin φ = 2/(3√6): φ = 15.8°M1 A1

Practice exercise 19

1. Find the equation of the plane containing L and perpendicular to Π.

21 Additional exam-style practice

Paper 1 and Paper 2.

WORKED EXAMPLE 20 Paper 2 question (exam level)

A(1, 0, 2), B(3, 1, 0), C(0, 2, 1), D(2, 3, 4). (a) Find the plane ABC. (b) Find the distance from D to ABC. (c) Find the volume of tetrahedron ABCD.

Solution

(a) AB × AC = (2, 1, −2) × (−1, 2, −1) = (3, 4, 5): 3x + 4y + 5z = 13M1 A1

(b) |6 + 12 + 20 − 13|/√50 = 25/√50 = 3.54M1 A1

Practice exercise 20

1. Find the area of triangle ABC above.

22 Exam-style questions with markscheme

Attempt each question under timed conditions before reading the markscheme.

Question 1 (Paper 1 style) (about 20 minutes)

Points A(1, 2, 0), B(3, 0, 1) and C(0, 1, 2).

(a) Find AB × AC. [3]

(b) Find the area of triangle ABC. [2]

(d) The line L passes through D(5, 5, 5) and is perpendicular to Π. Find the point where L meets Π. [4]

Markscheme

Part	Answer	Marks
(a)	AB = (2, −2, 1), AC = (−1, −1, 2)	A1
	(−4 + 1, −1 − 4, −2 − 2) = (−3, −5, −4)	M1 A1
(b)	½√(9 + 25 + 16)	M1
	½√50 = 5√2/2	A1
(c)	3x + 5y + 4z = 3 + 10 + 0	M1
	3x + 5y + 4z = 13	A1
(d)	r = (5, 5, 5) + λ(3, 5, 4)	A1
	3(5 + 3λ) + 5(5 + 5λ) + 4(5 + 4λ) = 13	M1
	60 + 50λ = 13, λ = −47/50	A1
	(2.18, 0.3, 1.24)	A1

Examiner's comments. Using n = (3, 5, 4) (a multiple of the cross product) simplifies the arithmetic.

Question 2 (Paper 2 style) (about 10 minutes)

Π1: x + 2y − z = 4 and Π2: 2x − y + z = 1.

(a) Find the acute angle between Π1 and Π2. [3]

(b) Find a vector equation of their line of intersection. [4]

Markscheme

Part	Answer	Marks
(a)	n1·n2 = 2 − 2 − 1 = −1	A1
	cos θ = 1/6	M1
	80.4°	A1
(b)	n1 × n2 = (1, −3, −5)	M1 A1
	z = 0: x + 2y = 4, 2x − y = 1: x = 1.2, y = 1.4	A1
	r = (1.2, 1.4, 0) + λ(1, −3, −5)	A1

Examiner's comments. Any correct point and any multiple of the direction is accepted.

23 Common misconceptions and links

Students often think…	Actually…
v × w = w × v	w × v = −v × w
the cross product is a number	it is a vector
the angle with a plane uses the normal directly	subtract from 90° or use sine
two planes always meet in a line	unless they are parallel

How 3.16 connects to the rest of the course

Topic	Connection
3.12 Vectors	scalar product
3.14 Lines	intersections
1.16 Systems	three planes
3.1 Triangles	½ab sin C

24 Check your understanding

Section A: multiple choice

1. i × j =

A. k B. −k C. 0 D. 1

2. v × w is perpendicular to

A. v only B. neither C. both v and w D. v + w

3. If v × w = 0, v and w are

A. perpendicular B. parallel C. equal D. unit vectors

4. A normal to 3x − y + 2z = 5 is

A. (3, −1, 5) B. (1, 1, 1) C. (5, 0, 0) D. (3, −1, 2)

5. The area of a triangle with sides v and w is

A. |v·w| B. ½|v × w| C. |v × w| D. ½|v||w|

Section B: structured

6. Find the plane through (1, 1, 1) perpendicular to the line r = λ(2, −1, 3). [2]

7. Find the distance from (2, 0, 1) to the plane 2x + y − 2z = 8. [3]

Answers

1. A 2. C 3. B 4. D 5. B

6. 2x − y + 3z = 4.

7. |4 + 0 − 2 − 8|/3 = 2.

25 Summary and topic map

Idea	Remember
Vector product	perpendicular to v and w; \|v\|\|w\| sin θ
Areas	parallelogram \|v × w\|; triangle ½\|v × w\|
Plane forms	r = a + λb + μc; r·n = a·n; ax + by + cz = d
Normal from points	AB × AC
Line meets plane	substitute; b·n = 0 means parallel
Angles	planes via normals; line-plane via sine
Two planes	line with direction n1 × n2
Distance	along the normal

Before the exam, make sure you…

•check normals with dot products
•use the simplest multiple of a normal
•subtract from 90° for line-plane angles
•find a point on the intersection line by setting a coordinate to zero

Vector Product and Planes on one page

Cover this page with a sheet of paper and try to rebuild it from memory. Whatever you can't recall is what to revise next.

Figure 2. Topic map for Vector Product and Planes.

26 Answers to practice exercises

Check your working against these answers. Full methods are expected in the exam.

Practice exercise 1

1. k; −k

2. (2, 5, −1)

3. (1, −1, 2)

Practice exercise 2

1. |(8, −4, −6)| = √116 = 10.8

2. ½|(−4, 8, −4)| = ½√96 = 4.90

Practice exercise 3

1. (2, −1, 3)

2. 3x + 4z = −1

3. 2 + 2 − 1 = 3: yes

Practice exercise 4

1. x − y + z = 0

2. 2x − y + z = 6

3. directions (1, 1, 1) and (−1, 1, 0): n = (−1, −1, 2): x + y − 2z = 1

Practice exercise 5

1. λ = 2: (2, 3, 4)

2. (1, −1, 0)·(1, 1, 1) = 0; (1, 1, 1) gives 3 ≠ 5

Practice exercise 6

1. cos θ = 4/(√3 √14): 51.9°

2. sin φ = 1/√3: 35.3°

Practice exercise 7

1. direction (1, −1, 0) × (0, 1, 1) = (−1, −1, 1); point (1, 0, 2): r = (1, 0, 2) + λ(−1, −1, 1)

2. the first two are parallel; the third cuts both, so no common point

Practice exercise 8

1. 3

2. λ = 1: (2, 2, 2)

Practice exercise 9

1. (−3, 6, −3)

2. (3, −2, 6)

Practice exercise 10

1. ½|(6, −3, 1)| = 3.39

2. (1, 2, 3)·((2, 1, 0) × (3, 3, 3)) = (1, 2, 3)·(3, −6, 3) = 0

Practice exercise 11

1. e.g. r = (2, 0, 0) + λ(1, −2, 0) + μ(0, 1, 1)

2. (4, 0, 0), (0, −6, 0), (0, 0, 2)

Practice exercise 12

1. x + y + z = 1

2. x − y + z = 0

Practice exercise 13

1. 2(2 + λ) − 2λ + 1 − λ = 9: λ = −4: (−2, −8, 5)

2. λ + λ − 1 − 2λ = −1 ≠ 5: parallel, no intersection

Practice exercise 14

1. 45°

2. 45°

Practice exercise 15

1. 3

Practice exercise 16

1. r = (4, 0, 0) + λ(1, −1, 0)

Practice exercise 17

1. (2, 2, 2)

Practice exercise 18

1. F = (2, 1, −2): image (4, 2, −4)

Practice exercise 19

1. n = (2, 1, −1) × (1, −2, 2) = (0, −5, −5) ∝ (0, 1, 1): y + z = 2

Practice exercise 20

1. ½√50 = 3.54

DP MATHEMATICS AA · TOPIC 1: NUMBER AND ALGEBRA

1.16

Systems of Linear Equations

Full notes for HL · Syllabus AHL 1.16

Contents

1	Systems of linear equations
2	Solving 2 × 2 systems
3	Solving 3 × 3 systems by elimination
4	Augmented matrices and row operations
5	Using technology
6	Unique, infinite and no solutions
7	Systems with a parameter
8	Geometric interpretation
9	Applications
10	Two equations in two unknowns: review
11	Three equations: elimination step by step
12	Row reduction with augmented matrices
13	Systems with infinitely many solutions
14	Inconsistent systems
15	Systems with a parameter
16	Geometric interpretation of three planes
17	Using technology to solve systems
18	Modelling with systems of equations
19	Mixed review: worked problems
20	Additional exam-style practice
21	Exam-style questions with markscheme
22	Common misconceptions and links
23	Check your understanding
24	Summary and topic map
25	Answers to practice exercises

These are full teaching notes. Read each section in order, try the Think about it question before you read on, and work through every worked example before attempting the Quick check.

Before you start

Rate yourself on each statement now, and again after you've worked through the notes. Shade R (not yet), A (getting there) or G (confident).

I can…	R	A	G
solve systems of two and three linear equations by substitution and elimination
use row operations on an augmented matrix to reach row echelon form
use technology to solve systems
recognize systems with a unique solution, infinitely many solutions or no solution
express infinite solutions in terms of a parameter
find values of a parameter for which a system is consistent
interpret systems geometrically as lines and planes
set up systems from real problems

This booklet covers Additional Higher Level (AHL) content (AHL 1.16). It is studied by HL students only.

Guiding questions

How can several equations be solved at once? What does it mean, algebraically and geometrically, when a system has no solution or infinitely many?

How this topic is examined

Systems are tested on Paper 1 (elimination, usually with a parameter) and Paper 2 (technology and modelling). They link directly to the intersection of planes in 3.16.

Show the row operations you use. When asked for the general solution, express every variable in terms of one parameter, such as λ.

Command term	What the examiner wants
Write down	an answer with little or no working; it should be obtainable by inspection
Find / Calculate / Determine	obtain the answer showing relevant working
Solve	find the values of the unknowns, using algebra or technology as allowed
Show that	obtain the given result; every step must be shown and the result must not be assumed
Hence / Hence or otherwise	use the previous result (or another valid method)
Sketch	a diagram showing the general shape and key features, labelled
Prove	a formal, rigorous argument from accepted facts
Justify / Explain	give valid reasons or evidence

How to use these notes

Each section teaches one part of the syllabus. You'll see these boxes throughout; each has one job:

Box	What it's for
Think about it	a question to get you thinking before a section
Why it works	a derivation or proof of a key result, so you understand where formulas come from
Key point	a result or idea you must not forget
Common mistake	an error examiners see every year, and how to avoid it
GDC tip	how to use your graphic display calculator efficiently on Paper 2
In the real world	where the mathematics is used outside the classroom
Worked example	a full solution with IB-style marks (M1 method, A1 answer, R1 reasoning, AG answer given); examples are graded basic, standard and exam level
Practice exercise	graded questions for each section; full answers are at the end of the booklet
Quick check	short questions to check the key ideas

1 Systems of linear equations

Several conditions, one solution set.

Think about it

You buy 2 pens and 3 notebooks for $12, and 4 pens and 1 notebook for $10. How can you find the price of each? What if the second shop had sold 4 pens and 6 notebooks for $24?

A linear equation has each unknown to the power 1, with no products of unknowns: 2x + 3y − z = 5 is linear; xy = 4 is not. A system of linear equations is a set of such equations, and a solution is a set of values satisfying every equation simultaneously.

A system is consistent if it has at least one solution, and inconsistent if it has none. The AA HL course covers systems of up to three equations in three unknowns.

2 Solving 2 × 2 systems

Substitution or elimination.

Substitution: make one variable the subject of one equation and substitute into the other. Elimination: multiply equations so that one variable has equal (or opposite) coefficients, then subtract (or add).

Geometrically, each equation is a straight line; the solution is the intersection point. Parallel lines (same gradient, different intercept) give no solution; identical lines give infinitely many.

WORKED EXAMPLE 1 Elimination (basic)

Solve 2p + 3n = 12 and 4p + n = 10.

Solution

Multiply the second by 3: 12p + 3n = 30M1

Subtract the first: 10p = 18, p = 1.8A1

n = 10 − 7.2 = 2.8: pens $1.80, notebooks $2.80A1

Practice exercise 1

1. Solve 3x − 2y = 7 and x + 4y = −7.

2. Solve 5x + 2y = 1 and 3x − y = 5.

3. For what value of k does kx + 2y = 3, 6x + 4y = 5 have no solution?

3 Solving 3 × 3 systems by elimination

Reduce to two equations, then one.

Use one equation to eliminate a variable from the other two, producing a 2 × 2 system. Solve it, then back-substitute to find the third variable. Choose the equation and variable that make the arithmetic easiest (a coefficient of 1 is ideal).

WORKED EXAMPLE 2 A 3 × 3 system (standard)

Solve x + y + z = 6, 2x − y + z = 3, x + 2y − z = 2.

Solution

(2) − 2(1): −3y − z = −9; (3) − (1): y − 2z = −4M1

From the second: y = 2z − 4; substitute: −3(2z − 4) − z = −9, so −7z = −21, z = 3A1

y = 2, x = 1: (x, y, z) = (1, 2, 3)A1

WORKED EXAMPLE 3 Fractions avoided (exam level)

Solve 2x + 3y − z = 5, x − y + 2z = 3, 3x + y + z = 10.

Solution

(1) − 2(2): 5y − 5z = −1; (3) − 3(2): 4y − 5z = 1M1

Subtract: y = −2; then z = −9/5A1

x = 3 + y − 2z = 3 − 2 + 18/5 = 23/5: (23/5, −2, −9/5)A1

Practice exercise 2

1. Solve x + y = 5, y + z = 7, x + z = 6.

2. Solve x + 2y + z = 8, 2x − y + z = 3, x + y − z = 0.

3. Solve 3x + y − 2z = 3, x − 2y + z = −2, 2x + 3y + z = 9.

4 Augmented matrices and row operations

A compact way to organize elimination.

A system can be written as an augmented matrix, with the coefficients on the left and the constants on the right. Three row operations do not change the solutions: swap two rows; multiply a row by a non-zero constant; add a multiple of one row to another.

Use row operations to reach row echelon form (a triangular arrangement with zeros below the leading entries), then back-substitute.

From the last row, −7z = −21 so z = 3; the second row gives y = 2; the first gives x = 1.

5 Using technology

The GDC simultaneous equation solver.

On Paper 2, a GDC equation solver or a matrix rref (reduced row echelon form) command solves systems quickly. Write down the system you entered and the solution.

GDC tip

If the solver reports 'infinite solutions' or 'no solution', use rref on the augmented matrix: a row of zeros gives infinitely many solutions; a row 0 0 0 | c with c ≠ 0 means no solution. You may still need to write the general solution by hand.

6 Unique, infinite and no solutions

Three possible outcomes.

After elimination, look at the final row. A row reading 0 = 0 (an equation that vanishes) means the system has infinitely many solutions: it is consistent, but one equation is a combination of the others. A row reading 0 = c with c ≠ 0 means no solution: the system is inconsistent. Otherwise there is a unique solution.

For infinitely many solutions, let one variable be a parameter, for example z = λ, and express the other variables in terms of λ. Each value of λ gives one particular solution.

WORKED EXAMPLE 4 Infinitely many solutions (standard)

Solve x + 2y − z = 1, 2x + y + z = 5, 4x + 5y − z = 7.

Solution

(2) − 2(1): −3y + 3z = 3; (3) − 4(1): −3y + 3z = 3M1

The two new equations are identical: one row reduces to 0 = 0, so infinitely many solutionsR1

Let z = λ: y = λ − 1, x = 1 − 2(λ − 1) + λ = 3 − λ: (3 − λ, λ − 1, λ)A1

WORKED EXAMPLE 5 No solution (standard)

Show that x + y + z = 1, 2x + y − z = 3, 3x + 2y = 7 has no solution.

Solution

(1) + (2): 3x + 2y = 4M1

But (3) says 3x + 2y = 7: 4 ≠ 7A1

The system is inconsistentR1

Practice exercise 3

1. Find the general solution of x + y + z = 3, x − y + 3z = 1, 2x + 4z = 4.

2. Show that x − y + z = 2, 2x + y − z = 1, 3x = 5 has no solution.

3. Determine whether x + 2y − z = 4, 2x + 4y − 2z = 9, x − y + z = 1 is consistent.

7 Systems with a parameter

Find when the system is consistent.

When a coefficient or constant is unknown (such as a or k), eliminate as usual and keep the parameter. The final equation has the form (expression in a)z = (expression in a, b). If the coefficient is non-zero, there is a unique solution. If it is zero, there are infinitely many solutions when the right side is also zero, and none otherwise.

WORKED EXAMPLE 6 Infinitely many or none (exam level)

Consider x + y + z = 2, x − y + 2z = 3, 2x + 4y + az = b. Find the values of a and b for which the system has (i) a unique solution, (ii) infinitely many solutions, (iii) no solution.

Solution

(2) − (1): −2y + z = 1; (3) − 2(1): 2y + (a − 2)z = b − 4M1

Add: (a − 1)z = b − 3A1

(i) a ≠ 1; (ii) a = 1 and b = 3; (iii) a = 1 and b ≠ 3A1 A1 A1

Practice exercise 4

1. Find k such that x + y = 3, 2x + ky = 6 has infinitely many solutions.

2. For x + y + z = 1, x + 2y + 3z = 2, x + 3y + kz = m, find k and m for infinitely many solutions.

8 Geometric interpretation

Each equation in three unknowns is a plane.

An equation ax + by + cz = d represents a plane in three dimensions (3.16). Three equations describe three planes, and the solution set is their common intersection.

A unique solution: the three planes meet at a single point. Infinitely many solutions: the planes meet in a line (a 'sheaf', like pages of a book meeting at the spine), or all three planes are the same. No solution: there is no common point; either two (or three) planes are parallel, or the planes form a triangular prism (each pair meets in a line, but the three lines are parallel).

Outcome	Final row after elimination	Geometric meaning
Unique solution	0 0 c \| d with c ≠ 0	planes meet at one point
Infinitely many	0 0 0 \| 0	planes meet in a line (sheaf) or coincide
No solution	0 0 0 \| d with d ≠ 0	parallel planes or a triangular prism

To distinguish a prism from parallel planes, compare the normal vectors (the coefficients): if no two planes have parallel normals but the system is inconsistent, the arrangement is a triangular prism.

9 Applications

Setting up systems from context.

WORKED EXAMPLE 7 Finding a quadratic (standard)

The curve y = ax2 + bx + c passes through (1, 2), (2, 7) and (−1, 4). Find a, b and c.

Solution

a + b + c = 2; 4a + 2b + c = 7; a − b + c = 4M1

(1) − (3): 2b = −2, b = −1; (2) − (1): 3a + b = 5, a = 2A1

c = 1: y = 2x2 − x + 1A1

WORKED EXAMPLE 8 A mixture problem (exam level)

A café sells small, medium and large coffees at $3, $4 and $5. One day it sells 120 coffees for $470, and twice as many small as large. How many of each were sold?

Solution

s + m + l = 120; 3s + 4m + 5l = 470; s = 2lM1

Substitute: 3l + m = 120 and 11l + 4m = 470M1

l = 10, s = 20, m = 90A1

Practice exercise 5

1. Find the quadratic through (0, 3), (1, 6) and (2, 13).

2. Three numbers sum to 20. The first is twice the second, and the third is 4 more than the second. Find them.

10 Two equations in two unknowns: review

Substitution and elimination.

For two linear equations in x and y, eliminate one variable by adding or subtracting multiples of the equations, or substitute one into the other. Geometrically, the solution is the intersection of two lines: one point, no point (parallel) or infinitely many (the same line).

WORKED EXAMPLE 9 Elimination (basic)

Solve 3x + 2y = 16 and 5x − 2y = 0.

Solution

Add: 8x = 16, x = 2M1 A1

x = 2, y = 5A1

Practice exercise 6

1. Solve 2x + 3y = 7 and 4x − y = 7.

2. For which k do 2x + ky = 3 and 4x + 6y = 1 have no solution?

11 Three equations: elimination step by step

Reduce to two, then one.

Use one equation to eliminate the same variable from the other two, giving two equations in two unknowns. Solve these, then back-substitute. Keep the work organized: label equations (1), (2), (3) and state each operation.

WORKED EXAMPLE 10 Three unknowns (standard)

Solve x + y + z = 6, 2x − y + z = 3, x + 2y − z = 2.

Solution

(1) + (3): 2x + 3y = 8; (2) + (3): 3x + y = 5M1 A1

Solve: x = 1, y = 2A1

x = 1, y = 2, z = 3A1

Practice exercise 7

1. Solve x + 2y − z = 3, 2x − y + z = 4, x + y + 2z = 5.

2. Solve 2x + y − z = 1, x − y + 2z = 5, 3x + 2y + z = 10.

12 Row reduction with augmented matrices

A systematic method.

Write the coefficients and constants as an augmented matrix and use row operations (swap rows, multiply a row by a non-zero constant, add a multiple of one row to another) to reach row-echelon form. Then back-substitute.

WORKED EXAMPLE 11 Row reduction (standard)

Solve x + y + z = 2, x + 2y + 3z = 5, 2x + 3y + 5z = 9.

Solution

R2 − R1: y + 2z = 3; R3 − 2R1: y + 3z = 5M1 A1

Subtract: z = 2, y = −1, x = 1A1

(1, −1, 2)A1

Practice exercise 8

1. Solve x + 2y + z = 4, 2x + 5y + 3z = 10, x + 3y + 4z = 8.

13 Systems with infinitely many solutions

A parameter describes the solutions.

If row reduction produces a row of zeros (0 = 0), one variable is free. Set it equal to a parameter λ and express the others in terms of λ. Geometrically, the three planes meet in a line.

WORKED EXAMPLE 12 Infinitely many (exam level)

Solve x + y + z = 3, x − y + 2z = 1, 2x + 3z = 4.

Solution

R1 + R2 = 2x + 3z = 4, same as R3: dependentM1 R1

Let z = λ: x = 2 − 1.5λ, y = 1 + 0.5λA1

Line of solutions (2 − 1.5λ, 1 + 0.5λ, λ)A1

Practice exercise 9

1. Find the general solution of x + 2y − z = 1, 2x + 4y − 2z = 2, x − y + z = 4.

14 Inconsistent systems

No solution.

If row reduction produces 0 = c with c ≠ 0, the system is inconsistent: the planes have no common point. They may be parallel, or form a triangular prism.

WORKED EXAMPLE 13 Inconsistent (standard)

Show that x + y + z = 1, x + y + z = 4, x − y = 0 has no solution.

Solution

R2 − R1: 0 = 3M1

Contradiction: no solution (first two planes are parallel)R1

Practice exercise 10

1. Show that x + 2y + z = 3, 2x + y − z = 1, 3x + 3y = 5 is inconsistent.

15 Systems with a parameter

For which values of k?

When a coefficient is an unknown k, reduce the system and look for values of k that make a pivot zero. For those values, decide whether the system has no solution or infinitely many by checking the constants.

WORKED EXAMPLE 14 Parameter (exam level)

Find k for which x + y + z = 1, 2x + 3y + kz = 3, x + 2y + 3z = 2 has no unique solution.

Solution

R2 − 2R1: y + (k − 2)z = 1; R3 − R1: y + 2z = 1M1 A1

Subtract: (k − 4)z = 0A1

k = 4: then 0 = 0, infinitely many solutionsA1 R1

Practice exercise 11

1. Find a such that x + y = 2, 2x + ay = 5 has no solution.

16 Geometric interpretation of three planes

Point, line, or none.

Comparing normal vectors (the coefficient rows) distinguishes the cases: equal or proportional normals mean parallel planes. If no two planes are parallel but there is no common point, they form a triangular prism.

Outcome	Geometry
unique solution	three planes meet at a point
infinitely many (one parameter)	planes meet in a line (sheaf)
infinitely many (two parameters)	all three planes coincide
no solution	parallel planes, or a triangular prism

WORKED EXAMPLE 15 Classifying (exam level)

Describe the intersection of x + y + z = 1, x + y + z = 2 and x − y = 0.

Solution

First two planes parallel and distinctR1

No common pointA1

17 Using technology to solve systems

GDC solvers and matrices.

On Paper 2, use the GDC simultaneous equation solver or the rref function on the augmented matrix. The rref form shows directly whether there is a unique solution, infinitely many (a zero row) or none (a row 0 0 0 | 1).

WORKED EXAMPLE 16 GDC (standard)

Solve 2x + 3y − z = 5, x − y + 4z = 11, 3x + 2y + z = 10 with technology.

Solution

rref gives x = 1, y = 2, z = 3A2

GDC tip

Enter the augmented matrix [A], then use rref([A]). A row [0 0 0 | 1] means no solution; a row of zeros means infinitely many.

18 Modelling with systems of equations

Mixtures, prices and fitting curves.

Many problems lead to linear systems: finding a quadratic through three points, mixing solutions of different concentrations, or finding prices from several purchases. Define variables clearly before writing equations.

WORKED EXAMPLE 17 Quadratic through three points (standard)

Find y = ax2 + bx + c through (1, 2), (2, 7) and (−1, 4).

Solution

a + b + c = 2; 4a + 2b + c = 7; a − b + c = 4M1 A1

Solve: a = 2, b = −1, c = 1A1

y = 2x2 − x + 1A1

WORKED EXAMPLE 18 Purchases (standard)

3 pens, 2 books and 1 bag cost $35; 1 pen, 3 books and 2 bags cost $43; 2 pens, 1 book and 3 bags cost $42. Find each price.

Solution

Solve the systemM1

Pen $4, book $7, bag $9A1

Practice exercise 12

1. Find the circle x2 + y2 + Dx + Ey + F = 0 through (0, 0), (4, 0) and (0, 6).

19 Mixed review: worked problems

Bringing the ideas together.

WORKED EXAMPLE 19 Mixed problem (exam level)

Solve x + 2y + 3z = 14, 2x + y + z = 7, 3x − y + 2z = 7.

Solution

Eliminate to two equations, solveM1 A1

x = 1, y = 2, z = 3A1

Practice exercise 13

1. Solve x − y + z = 2, 3x + y − 2z = −1, 2x + 2y + z = 9.

20 Additional exam-style practice

Paper 1 and Paper 2.

WORKED EXAMPLE 20 Paper 1 question (exam level)

Consider x + y + z = 4, x + 2y + 3z = 7, 2x + 3y + az = b. (a) Find a for which the system does not have a unique solution. (b) For this a, find b for which there are infinitely many solutions.

Solution

(a) R3 − R1 − R2: (a − 4)z = b − 11: a = 4M1 A1

(b) b = 11A1

Practice exercise 14

1. For a = 4, b = 11, find the general solution.

21 Exam-style questions with markscheme

Attempt each question under timed conditions before reading the markscheme.

Question 1 (Paper 1 style) (about 15 minutes)

Consider the system: x + 2y + z = 3; 2x + y − z = 0; 4x + 5y + kz = 6.

(a) Show that the system can be reduced to 3y + 3z = 6 and 3y + (4 − k)z = 6. [3]

(b) Find the value of k for which the system has infinitely many solutions. [2]

(d) Describe the geometric arrangement of the three planes for this value of k. [1]

Markscheme

Part	Answer	Marks
(a)	2(1) − (2): 3y + 3z = 6	M1 A1
	4(1) − (3): 3y + (4 − k)z = 6	A1
(b)	subtract: (1 − k)z = 0	M1
	k = 1	A1
(c)	z = λ, y = 2 − λ	A1
	x = 3 − 2(2 − λ) − λ = λ − 1	A1
	(λ − 1, 2 − λ, λ)	A1
(d)	planes meet in a line (sheaf)	A1

Examiner's comments. Part (a) is a show-that: every row operation must be stated.

Question 2 (Paper 2 style) (about 10 minutes)

A company makes three products, A, B and C, using three machines. Machine hours per unit: machine 1: A 2, B 1, C 3; machine 2: A 1, B 2, C 1; machine 3: A 3, B 1, C 2. Available hours: 180, 120 and 180.

(a) Write a system of equations for using all available hours. [2]

(b) Find the number of units of each product. [2]

Markscheme

Part	Answer	Marks
(a)	2a + b + 3c = 180; a + 2b + c = 120; 3a + b + 2c = 180	A2
(b)	GDC solver	M1
	a = 30, b = 30, c = 30	A1

Examiner's comments. Clear definition of variables earns credit in (a).

22 Common misconceptions and links

Students often think…	Actually…
0 = 0 means no solution	0 = 0 means infinitely many
giving one particular solution when asked for the general solution	use a parameter
dividing by an expression that could be zero	consider that case separately
no solution always means parallel planes	it could be a triangular prism

How 1.16 connects to the rest of the course

Topic	Connection
3.16 Planes	equations and intersections of planes
1.11 Partial fractions	comparing coefficients
2.12 Polynomials	finding coefficients
2.1 Lines	intersection of lines in 2D

23 Check your understanding

Section A: multiple choice

1. A row 0 0 0 | 4 means

A. no solution B. infinitely many C. a unique solution D. x = 4

2. Three planes meeting at a point give

A. no solution B. infinitely many C. a unique solution D. a line

3. A row 0 0 0 | 0 means

A. no solution B. infinitely many solutions C. a unique solution D. an error

Section B: structured

4. Solve x + y = 5, y + z = 7, x + z = 6. [3]

5. Describe a triangular prism arrangement of planes. [2]

Answers

1. A 2. C 3. B

4. x = 2, y = 3, z = 4.

5. Each pair of planes meets in a line; the three lines are parallel; there is no common point.

24 Summary and topic map

Idea	Remember
Elimination	reduce to triangular form; back-substitute
Row operations	swap, scale, add multiples
Technology	solver or rref on Paper 2
Outcomes	unique; infinite (0 = 0); none (0 = c)
General solution	parameter λ
Geometry	point, line, prism, parallel planes

Before the exam, make sure you…

•state row operations
•identify 0 = 0 and 0 = c rows
•use a parameter for infinite solutions
•link to planes

Systems of Linear Equations on one page

Cover this page with a sheet of paper and try to rebuild it from memory. Whatever you can't recall is what to revise next.

Figure 1. Topic map for Systems of Linear Equations.

25 Answers to practice exercises

Check your working against these answers. Full methods are expected in the exam.

Practice exercise 1

1. x = 1, y = −2

2. x = 1, y = −2

3. k = 3 (parallel lines)

Practice exercise 2

1. x = 2, y = 3, z = 4

2. x = 1, y = 2, z = 3

3. x = 1, y = 2, z = 1

Practice exercise 3

1. (2 − 2λ, 1 + λ, λ)

2. (1) + (2) gives 3x = 3, which contradicts 3x = 5

3. no: 2 × (1) gives 8, not 9

Practice exercise 4

1. k = 2

2. row reduce: (k − 5)z = m − 3: k = 5, m = 3

Practice exercise 5

1. y = 2x2 + x + 3

2. 8, 4, 8

Practice exercise 6

1. x = 2, y = 1

2. k = 3

Practice exercise 7

1. x = 2, y = 1, z = 1

2. x = 1, y = 2, z = 3

Practice exercise 8

1. x = 1, y = 1, z = 1

Practice exercise 9

1. z = λ: x = 3 − λ/3, y = 2λ/3 − 1

Practice exercise 10

1. R1 + R2: 3x + 3y = 4 ≠ 5

Practice exercise 11

1. a = 2

Practice exercise 12

1. F = 0, D = −4, E = −6

Practice exercise 13

1. x = 1, y = 2, z = 3

Practice exercise 14

1. z = λ: y = 3 − 2λ, x = 1 + λ

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