Mini-Lesson

Vectors

3 missions. Guess, learn, then check. Your place is saved.

Mission 1How can one number tell you if two directions are at right angles?
Mission 2Two aircraft fly straight paths. Will they collide or just pass close?
Mission 3Three points fix a plane. Which direction sticks straight out of it?

The specification points and sources are listed under the lesson: see Sources.

Mission 1

How can one number tell you if two directions are at right angles?

Make a guess: How can one number tell you if two directions are at right angles?

Vectors · scalar product

Components, magnitude and unit vectors

A vector has size and direction: $v = v_{1} v_{2} v_{3} = v_{1} i + v_{2} j + v_{3} k$ v = ( v₁ ; v₂ ; v₃ ) = v₁i + v₂j + v₃k.
Position vector $O A = a$ OA→ = a. Displacement $A B = b - a$ AB→ = b − a (end minus start).
Magnitude: $∣ v ∣ = v_{12} + v_{22} + v_{32}$ |v| = √(v₁² + v₂² + v₃²), so the distance $A B$ AB is $∣ A B ∣$ |AB→|.
A unit vector has magnitude 1: $\frac{v}{∣ v ∣}$ v/(|v|). Vectors are parallel if one is a scalar multiple of the other, $w = k v$ w = kv.

Calculate

?Points

A

and

B

have coordinates

(1, - 2, 3)

and

(3, 4, 0)

. Find the distance

A B

.

Worked example Vectors · scalar product

A velocity with a given speed

Question

A drone flies at a constant speed of 15 m s⁻¹ in the direction of

2 i - 2 j + k

. Find its velocity vector.

1
Find the magnitude of the direction vector: $2^{2} + (- 2)^{2} + 1^{2} = 9 = 3$ √(2² + (−2)² + 1²) = √9 = 3
2
Normalise to get a unit vector: $\frac{1}{3} 2 - 2 1$ 1/3( 2 ; −2 ; 1 )
3
Rescale to the speed: $15 \times \frac{1}{3} 2 - 2 1 = 10 - 10 5$ 15 × 1/3( 2 ; −2 ; 1 ) = ( 10 ; −10 ; 5 )

Answer

velocity

= 10 i - 10 j + 5 k

m s⁻¹. Check:

100 + 100 + 25 = 15

.

⚠️ Watch out

15 (2 i - 2 j + k)

has magnitude 45, not 15. Normalise first.

Vectors · scalar product

The scalar product

The scalar product (dot product) of $v$ v and $w$ w is a number, not a vector:

v \cdot w = v_{1} w_{1} + v_{2} w_{2} + v_{3} w_{3} = ∣ v ∣∣ w ∣ cos θ

$θ$ θ is the angle between the vectors when they point away from the same point. So $cos θ = \frac{v \cdot w}{∣ v ∣∣ w ∣}$ cos θ = (v·w)/(|v||w|).
For non-zero vectors: $v \cdot w > 0$ v·w > 0 means $θ$ θ is acute, $v \cdot w = 0$ v·w = 0 means perpendicular, and $v \cdot w < 0$ v·w < 0 means obtuse.
Properties: $v \cdot w = w \cdot v$ v·w = w·v, $u \cdot (v + w) = u \cdot v + u \cdot w$ u·(v + w) = u·v + u·w, $(k v) \cdot w = k (v \cdot w)$ (kv)·w = k(v·w) and $v \cdot v = ∣ v ∣^{2}$ v·v = |v|². For parallel vectors, $∣ v \cdot w ∣ = ∣ v ∣∣ w ∣$ |v·w| = |v||w|. You can use these in vector proofs.

Worked example Vectors · scalar product

The angle between two vectors

Question

Find the angle between

p = 21 - 2

and

q = 304

, in degrees.

1
Scalar product: $p \cdot q = 6 + 0 - 8 = - 2$ p·q = 6 + 0 − 8 = −2
2
Magnitudes: $∣ p ∣ = 4 + 1 + 4 = 3$ |p| = √(4 + 1 + 4) = 3 and $∣ q ∣ = 9 + 0 + 16 = 5$ |q| = √(9 + 0 + 16) = 5
3
Cosine: $cos θ = \frac{- 2}{3 \times 5} = - \frac{2}{15}$ cos θ = (−2)/(3 × 5) = −2/15
4
Inverse cosine: $θ = arccos (- \frac{2}{15})$ θ = arccos (−2/15)

Answer

θ \approx 97. 7^{\circ}

(3 s.f.), an obtuse angle.

⚠️ Watch out

keep the minus sign. Dropping it gives

82. 3^{\circ}

, the supplement of the true angle.

Calculate

?Points

A (1, 0, 2)

,

B (3, 1, 0)

and

C (4, k, 5)

are such that angle

A B^C = 9 0^{\circ}

. Find the value of

k

.

Worked example Vectors · scalar product

A vector proof: the angle in a semicircle

Question

A B

is a diameter of a circle with centre

O

, and

C

is another point on the circle. Let

O A = a

and

OC = c

. Prove that

A C^B = 9 0^{\circ}

.

1
Use the circle: $O$ O is the midpoint of $A B$ AB, so $OB = - a$ OB→ = −a; radii are equal, so $∣ c ∣ = ∣ a ∣$ |c| = |a|
2
Write the sides from $C$ C: $C A = a - c$ CA→ = a − c and $CB = - a - c$ CB→ = −a − c
3
Expand the scalar product: $C A \cdot CB = - (a - c) \cdot (a + c) = - (a \cdot a + a \cdot c - c \cdot a - c \cdot c)$ CA→·CB→ = −(a − c)·(a + c) = −(a·a + a·c − c·a − c·c)
4
Simplify with $a \cdot c = c \cdot a$ a·c = c·a and $v \cdot v = ∣ v ∣^{2}$ v·v = |v|²: $C A \cdot CB = - (∣ a ∣^{2} - ∣ c ∣^{2}) = 0$ CA→·CB→ = −(|a|² − |c|²) = 0
5
Conclude: $C$ C is not $A$ A or $B$ B, so both vectors are non-zero, and a zero scalar product means they are perpendicular

Answer

C A \cdot CB = 0

, so

A C^B = 9 0^{\circ}

, as required.

⚠️ Watch out

justify each fact you use (equal radii,

OB = - a

, non-zero vectors). Unexplained steps lose marks in a "prove" question.

Sort it

?Use the scalar product (and scalar multiples) to classify each pair of vectors.

Tap a statement, then tap the group it belongs in. Then press Check.

Mission check

?How can one number tell you if two directions are at right angles? Take

u = 2 - 3 1

and

v = 42 - 2

.

Mission 2

Two aircraft fly straight paths. Will they collide or just pass close?

Make a guess: Two aircraft fly straight paths. Will they collide or just pass close?

Vectors · lines

The vector equation of a line

A line through the point with position vector $a$ a, in the direction of $b$ b, is

r = a + λ b, λ \in R

$a$ a fixes where the line is; the direction vector $b$ b fixes which way it points. Each value of $λ$ λ gives one point on the line.
Parametric form: write each component separately, $x = x_{0} + λ l$ x = x₀ + λ l, $y = y_{0} + λm$ y = y₀ + λ m, $z = z_{0} + λn$ z = z₀ + λ n.
Cartesian form: make $λ$ λ the subject of each and equate, $\frac{x - x _{0}}{l} = \frac{y - y _{0}}{m} = \frac{z - z _{0}}{n}$ (x − x₀)/l = (y − y₀)/m = (z − z₀)/n.

Quick check

?Which point lies on the line

r = 12 - 1 + λ 2 - 1 3

?

Vectors · lines

Parallel, intersecting or skew

Compare the direction vectors first. If one is a multiple of the other, the lines are parallel (or coincident, if they share a point).
If not, set the two equations equal: $a_{1} + λ b_{1} = a_{2} + μ b_{2}$ a₁ + λb₁ = a₂ + μb₂. Use a different letter for each parameter.
Solve two of the component equations for $λ$ λ and $μ$ μ, then test the third.
Third equation true: the lines intersect. False: in 3-D the lines are skew; they are not parallel and never meet.

Worked example Vectors · lines

Finding a point of intersection

Question

Show that the lines

r = 1 - 1 2 + λ 21 - 1

and

r = 3 - 3 2 + μ 12 - 1

intersect, and find the point of intersection.

1
Direction vectors are not multiples of each other, so the lines are not parallel
2
Equate $x$ x and $y$ y: $1 + 2 λ = 3 + μ$ 1 + 2λ = 3 + μ and $- 1 + λ = - 3 + 2 μ$ −1 + λ = −3 + 2μ
3
Substitute $μ = 2 λ - 2$ μ = 2λ − 2 into the second: $- 1 + λ = - 7 + 4 λ$ −1 + λ = −7 + 4λ, so $λ = 2$ λ = 2 and $μ = 2$ μ = 2
4
Test $z$ z: $2 - λ = 0$ 2 − λ = 0 and $2 - μ = 0$ 2 − μ = 0. Both are 0, so the lines intersect
5
Substitute $λ = 2$ λ = 2 into the first line: $(1 + 4, - 1 + 2, 2 - 2)$ (1 + 4, −1 + 2, 2 − 2)

Answer

the lines intersect at

(5, 1, 0)

.

⚠️ Watch out

you must test the third equation. If the second line started at

(3, - 3, 5)

, the same

λ

and

μ

would fail the

z

test and the lines would be skew.

Calculate

?Find the acute angle between the lines

r = 014 + λ 122

and

r = 30 - 1 + μ 2 - 1 - 2

. Give your answer in degrees to 3 significant figures.

Vectors · lines

Lines as motion

An object moving with constant velocity $v$ v from position $r_{0}$ r₀ has position $r = r_{0} + t v$ r = r₀ + tv at time $t$ t. The parameter is now time.
The speed is $∣ v ∣$ |v|. The relative position of $B$ B from $A$ A is $A B = r_{B} - r_{A}$ AB→ = r_B − r_A.
Two paths can cross with no collision. A collision needs the same point at the same time: solve $r_{A} (t) = r_{B} (t)$ r_A(t) = r_B(t) with one $t$ t.

Worked example Vectors · lines

Crossing paths

Question

Positions in km relative to a control tower,

t

minutes after noon, are

r_{A} = 123 + t 210

and

r_{B} = 8 - 2 0 + t - 1 21

. Do the paths cross, and do the aircraft collide?

1
Paths: use different parameters, $1 + 2 s = 8 - u$ 1 + 2s = 8 − u and $2 + s = - 2 + 2 u$ 2 + s = −2 + 2u, giving $s = 2$ s = 2, $u = 3$ u = 3
2
Test $z$ z: $3 = 0 + 3$ 3 = 0 + 3, true. The paths cross at $(5, 4, 3)$ (5, 4, 3)
3
Times: $A$ A is there when $t = 2$ t = 2, but $B$ B only when $t = 3$ t = 3
4
Collision test: $x$ x needs $1 + 2 t = 8 - t$ 1 + 2t = 8 − t, so $t = \frac{7}{3}$ t = 7/3, but $z$ z needs $3 = t$ 3 = t. No single $t$ t works

Answer

the paths cross at

(5, 4, 3)

, but the aircraft do not collide:

B

arrives 1 minute after

A

has passed.

⚠️ Watch out

using the same letter for both parameters in step 1 would wrongly suggest the paths never meet.

Calculate

?Two drones move with positions

r_{P} = 120 + t 2 - 1 1

and

r_{Q} = 7 - 4 9 + t 01 - 2

, in metres,

t

seconds after launch. They collide. Find the value of

t

when they collide.

Mission check

?Two aircraft fly straight paths. Will they collide or just pass close? Take

r_{A} = 012 + t 110

and

r_{B} = 100 + t 011

.

Mission 3

Three points fix a plane. Which direction sticks straight out of it?

Make a guess: Three points fix a plane. Which direction sticks straight out of it?

Vectors · planes

The vector product

The vector product (cross product) is a vector perpendicular to both $v$ v and $w$ w:

v \times w = v_{2} w_{3} - v_{3} w_{2} v_{3} w_{1} - v_{1} w_{3} v_{1} w_{2} - v_{2} w_{1}

$v \times w = ∣ v ∣∣ w ∣ sin θ n$ v×w = |v||w|sin θ n, where $n$ n is a unit normal set by the right-hand screw rule.
Order matters: $w \times v = - v \times w$ w×v = −v×w. Also $v \times v = 0$ v×v = 0, and for non-zero vectors $v \times w = 0$ v×w = 0 means parallel.
$∣ v \times w ∣$ |v×w| is the area of the parallelogram with sides $v$ v and $w$ w; a triangle is half.

Quick check

?Find

2 - 1 3 \times 14 - 2

.

Vectors · planes

Equations of a plane

Vector form: $r = a + λ b + μ c$ r = a + λb + μc, where $a$ a is a point and $b$ b, $c$ c are non-parallel vectors in the plane.
A normal $n$ n is perpendicular to every vector in the plane. You can take $n = b \times c$ n = b×c.
Scalar product form: $r \cdot n = a \cdot n$ r·n = a·n.
Cartesian form: with $n = n_{1} n_{2} n_{3}$ n = ( n₁ ; n₂ ; n₃ ), this is $n_{1} x + n_{2} y + n_{3} z = d$ n₁x + n₂y + n₃z = d, where $d = a \cdot n$ d = a·n.

Worked example Vectors · planes

The plane through three points

Question

Find a Cartesian equation of the plane through

A (1, 0, 2)

,

B (3, 1, 0)

and

C (2, - 1, 4)

.

1
Two vectors in the plane: $A B = 21 - 2$ AB→ = ( 2 ; 1 ; −2 ) and $A C = 1 - 1 2$ AC→ = ( 1 ; −1 ; 2 )
2
Normal: $A B \times A C = 2 - 2 - 2 - 4 - 2 - 1 = 0 - 6 - 3$ AB→×AC→ = ( 2 − 2 ; −2 − 4 ; −2 − 1 ) = ( 0 ; −6 ; −3 ), so use $n = 021$ n = ( 0 ; 2 ; 1 )
3
Find $d$ d using $A$ A: $a \cdot n = 0 + 0 + 2 = 2$ a·n = 0 + 0 + 2 = 2
4
Check with $B$ B and $C$ C: $2 (1) + 0 = 2$ 2(1) + 0 = 2 and $2 (- 1) + 4 = 2$ 2(−1) + 4 = 2

Answer

2 y + z = 2

⚠️ Watch out

any non-zero multiple of a normal is also a normal. Dividing by

- 3

keeps the numbers small.

Calculate

?Find the area of triangle

A BC

from the previous example, to 3 significant figures.

Vectors · planes

Intersections and angles

Line and plane: substitute the line's $x$ x, $y$ y, $z$ z (in terms of $λ$ λ) into the plane's equation and solve. One $λ$ λ: one point. No solution: parallel. True for every $λ$ λ: the line lies in the plane.
Two planes: non-parallel planes meet in a line, with direction $n_{1} \times n_{2}$ n₁×n₂. Three planes: solve the system (a point, a line, a plane, or no common point).
Angles: between two planes, the acute angle has $cos θ = \frac{∣ n _{1} \cdot n _{2} ∣}{∣ n _{1} ∣∣ n _{2} ∣}$ cos θ = (|n₁·n₂|)/(|n₁||n₂|). Between a line and a plane, $sin θ = \frac{∣ b \cdot n ∣}{∣ b ∣∣ n ∣}$ sin θ = (|b·n|)/(|b||n|).

Worked example Vectors · planes

Where a line meets a plane

Question

Find the point where the line

r = 120 + λ 1 - 1 2

meets the plane

2 x + y - z = 3

.

1
Parametric form: $x = 1 + λ$ x = 1 + λ, $y = 2 - λ$ y = 2 − λ, $z = 2 λ$ z = 2λ
2
Substitute: $2 (1 + λ) + (2 - λ) - 2 λ = 3$ 2(1 + λ) + (2 − λ) − 2λ = 3
3
Simplify: $4 - λ = 3$ 4 − λ = 3, so $λ = 1$ λ = 1
4
Put $λ = 1$ λ = 1 into the line: $(1 + 1, 2 - 1, 2)$ (1 + 1, 2 − 1, 2)

Answer

the line meets the plane at

(2, 1, 2)

. Check:

4 + 1 - 2 = 3

.

⚠️ Watch out

substitute the line into the plane, not the plane's normal into the line.

Calculate

?Find the acute angle between the line and the plane in the previous example, in degrees to 3 significant figures.

Sort it

?What is the intersection in each case?

Tap a statement, then tap the group it belongs in. Then press Check.

Mission check

?Three points fix a plane. Which direction sticks straight out of it? The plane contains

P (1, 0, 0)

,

Q (0, 2, 0)

and

R (0, 0, 3)

.

Recap

Big ideas

$A B = b - a$ AB→ = b − a, $∣ v ∣ = v_{12} + v_{22} + v_{32}$ |v| = √(v₁² + v₂² + v₃²), and a unit vector is $\frac{v}{∣ v ∣}$ v/(|v|): normalise, then rescale.
$v \cdot w = v_{1} w_{1} + v_{2} w_{2} + v_{3} w_{3} = ∣ v ∣∣ w ∣ cos θ$ v·w = v₁w₁ + v₂w₂ + v₃w₃ = |v||w|cos θ gives the angle between vectors; zero means perpendicular.
A line is $r = a + λ b$ r = a + λb: a point plus multiples of a direction; the angle between lines comes from the direction vectors.
To compare two lines: parallel if the directions are multiples; otherwise solve with two parameters and test the third equation (intersect or skew).
With $r = r_{0} + t v$ r = r₀ + tv, a collision needs the same point at the same time; crossing paths alone don't collide.
$v \times w$ v×w is perpendicular to both vectors; $∣ v \times w ∣ = ∣ v ∣∣ w ∣ sin θ$ |v×w| = |v||w|sin θ is the parallelogram's area.
A plane is $r = a + λ b + μ c$ r = a + λb + μc, or $r \cdot n = a \cdot n$ r·n = a·n, or $n_{1} x + n_{2} y + n_{3} z = d$ n₁x + n₂y + n₃z = d, with normal $n = b \times c$ n = b×c.
Substitute a line into a plane to intersect them; use normals for angles between planes, and $sin θ = \frac{∣ b \cdot n ∣}{∣ b ∣∣ n ∣}$ sin θ = (|b·n|)/(|b||n|) for a line and a plane.

🏆

Mini-lesson complete!

⭐⭐⭐

You've worked through Vectors for IB Diploma Mathematics: Analysis & Approaches HL. 🎉

Mission 1How can one number tell you if two directions are at right angles?
Mission 2Two aircraft fly straight paths. Will they collide or just pass close?
Mission 3Three points fix a plane. Which direction sticks straight out of it?

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Specification points and sources for this lesson are listed at the bottom of the page.