Vectors: A-level Maths OCR revision notes

1.10

Vectors in two and three dimensions

Vector notation and components 1.10

Definitions

Vector: a quantity with both magnitude and direction, such as displacement, velocity or force.
Scalar: a quantity with magnitude only, such as distance, speed or mass.
Zero vector $0$ : the vector with all components zero; it has zero magnitude and no defined direction.

Key results

A vector can be written in column form $x y z$ or in terms of unit vectors as $x i + y j + z k$ ; both notations describe the same displacement, and questions move between them freely.
$i$ , $j$ and $k$ are the unit vectors of length $1$ along the positive $x$ -, $y$ - and $z$ -axes respectively.
Two vectors are equal exactly when all of their corresponding components are equal.

Notes

A vector carries no fixed starting point: the same components describe the same displacement wherever it is drawn, which is why $A B$ and a parallel, equal-length arrow elsewhere in the diagram are the same vector.
In handwriting a vector is underlined ( $\underline{a}$ ); in print it is bold ( $a$ ). Write $∣ a ∣$ , never $a$ , when you mean its length.
Column form and $i$ - $j$ - $k$ form contain exactly the same information — pick whichever the question uses, or whichever makes the arithmetic that follows easiest.

Magnitude and unit vectors 1.10

Definitions

Magnitude (modulus) $∣ v ∣$ : the length of the vector $v$ .
Unit vector: a vector of magnitude $1$ .

Key results

In two dimensions $∣ v ∣ = x^{2} + y^{2}$ ; in three dimensions $∣ v ∣ = x^{2} + y^{2} + z^{2}$ , the three-dimensional extension of Pythagoras.
The unit vector in the direction of $v$ is $\hat{v} = \frac{v}{∣ v ∣}$ .
Scaling behaves predictably: $∣ k v ∣ = ∣ k ∣ ∣ v ∣$ for any scalar $k$ .

Method

To construct a vector of stated length $λ$ in the direction of $v$ : compute $∣ v ∣$ , divide $v$ by it to get the unit vector, then multiply by $λ$ , giving $\frac{λ}{∣ v ∣} v$ .

In practiceFind a vector of length 15 in the direction of $v = 2 i - j + 2 k$ .

$∣ v ∣ = 4 + 1 + 4 = 3$
$\hat{v} = \frac{1}{3} (2 i - j + 2 k)$ the unit vector
$15 \hat{v} = 10 i - 5 j + 10 k$ check: length $100 + 25 + 100 = 15$

Notes

A magnitude is never negative, and $∣ v ∣ = 0$ only for the zero vector.
The signs of the components are lost inside the squares, so a unit vector must be built from the original vector, not from its magnitude alone.
Dividing by $∣ v ∣$ to get $\hat{v}$ is only valid when $v \neq = 0$ — the zero vector has no direction, so it has no associated unit vector.

Magnitude-direction form in two dimensions 1.10

Key results

A two-dimensional vector $(x y)$ has magnitude $r = x^{2} + y^{2}$ and direction $θ$ satisfying $tan θ = \frac{y}{x}$ , usually measured anticlockwise from the positive $x$ -direction.
Converting back: $x = r cos θ$ and $y = r sin θ$ .
Elementary direction work of this kind is done in degrees, consistent with the rest of the course's non-calculus trigonometry.

Notes

A calculator's $tan^{- 1}$ returns an angle between $- 9 0^{\circ}$ and $9 0^{\circ}$ , so it gives the wrong answer for vectors pointing into the second or third quadrant — always sketch the vector and adjust by $18 0^{\circ}$ where necessary.
In mechanics and navigation the direction is often quoted instead as a bearing, measured clockwise from north as a three-figure angle.
Converting between the two forms and back should return the original vector — a quick way to check $r$ and $θ$ were both found correctly before using them further.

Worked example

The vector $a = 3 i - 4 j$ .

Find $∣ a ∣$ , the unit vector in the direction of $a$ , and the angle $a$ makes with the positive $x$ -direction.

Show worked solution

∣ a ∣ = 3^{2} + (- 4)^{2} = 9 + 16 = 25 = 5

The unit vector is:

\hat{a} = \frac{a}{∣ a ∣} = \frac{3}{5} i - \frac{4}{5} j = 0.6 i - 0.8 j

and as a check:

0. 6^{2} + 0. 8^{2} = 0.36 + 0.64 = 1

For the direction,

tan θ = \frac{- 4}{3}

and since the vector points right and down it lies in the fourth quadrant, so $θ = - 53. 1^{\circ}$ (3 s.f.), that is $53. 1^{\circ}$ below the positive $x$ -direction.

1.10

Vector arithmetic

Adding, subtracting and scaling vectors 1.10

Key results

Vectors add and subtract component-wise: $(a_{1} a_{2}) + (b_{1} b_{2}) = (a_{1} + b_{1} a_{2} + b_{2})$ .
Triangle law: placing $b$ nose-to-tail after $a$ gives the resultant $a + b$ ; equivalently $A B + BC = A C$ .
Subtraction is addition of the reverse vector: $a - b = a + (- b)$ , and $- b$ has the same length as $b$ but the opposite direction.
Multiplying a vector by a scalar changes its magnitude, and reverses its direction if the scalar is negative, without changing the line the vector lies along.

Notes

Vector addition is commutative and associative, which is what makes the parallelogram and triangle constructions give the same resultant.
$∣ a + b ∣$ is not generally $∣ a ∣ + ∣ b ∣$ — the two are equal only when $a$ and $b$ point in the same direction. Add the vectors first, then take the magnitude.
$- b$ reverses every component's sign, not just one — a common slip is negating only the first component when forming $a - b$ .

Parallel vectors 1.10

Key results

$a$ is parallel to $b$ exactly when $a = k b$ for some non-zero scalar $k$ .
If $k > 0$ the two vectors point in the same direction; if $k < 0$ they point in opposite directions but still lie along parallel lines.
Equivalently, corresponding components are in a constant ratio: $\frac{a _{1}}{b _{1}} = \frac{a _{2}}{b _{2}} = \frac{a _{3}}{b _{3}}$ .

Method

Divide each component of $a$ by the corresponding component of $b$ .
If every ratio gives the same value $k$ , the vectors are parallel and $a = k b$ ; if any ratio differs, they are not parallel.

In practiceIs $a = 6 i - 9 j + 3 k$ parallel to $b = - 2 i + 3 j - k$ ?

$\frac{6}{- 2} = - 3, \frac{- 9}{3} = - 3, \frac{3}{- 1} = - 3$ every ratio the same
$a = - 3 b$ parallel, pointing in opposite directions

Notes

Two vectors of equal magnitude need not be parallel, and two parallel vectors need not have equal magnitude — the two properties are independent.
If a component of $b$ is zero, that ratio can't be formed by division — check that component pairs directly instead (both zero, or neither) rather than dividing by zero.

Worked example

Given:

v = 2 i - 3 j + 6 k

find $∣ v ∣$ and the values of the scalar $λ$ for which $∣ λ v ∣ = 21$ .

Show worked solution

∣ v ∣ = 2^{2} + (- 3)^{2} + 6^{2} = 4 + 9 + 36 = 49 = 7

Since:

∣ λ v ∣ = ∣ λ ∣ ∣ v ∣ = 7∣ λ ∣

the requirement $7∣ λ ∣ = 21$ gives $∣ λ ∣ = 3$ , so $λ = 3$ or $λ = - 3$ .

Both work:

3 v = 6 i - 9 j + 18 k

has magnitude:

36 + 81 + 324 = 441 = 21

and $- 3 v$ has the same magnitude in the opposite direction.

1.10

Position vectors and geometric problems

Position vectors, displacement and distance 1.10

Definitions

Position vector of a point $A$ : the displacement $O A = a$ of that point from a fixed origin $O$ .

Key results

The vector from $A$ to $B$ is $A B = b - a$ , the position vector of $B$ minus the position vector of $A$ .
The distance between $A$ and $B$ is $∣ A B ∣ = (b_{1} - a_{1})^{2} + (b_{2} - a_{2})^{2} + (b_{3} - a_{3})^{2}$ , which is exactly the three-dimensional distance formula.
Reversing the direction reverses the sign: $B A = - A B$ , though the distance $∣ B A ∣ = ∣ A B ∣$ is unchanged.

Notes

Subtracting in the wrong order is the single most common vector error; read $A B$ as 'finish minus start'.
Points are written with coordinates, $A (1, 2, 3)$ ; their position vectors are written as vectors, $a = i + 2 j + 3 k$ . Keep the two notations distinct in a written solution.
The distance formula here is exactly the 2D distance formula from Coordinate geometry with one extra squared term — recognising the pattern saves re-deriving it from scratch.

Collinearity and the midpoint 1.10

Definitions

Collinear points: three or more points that all lie on a single straight line.

Key results

The midpoint $M$ of $A B$ has position vector $m = \frac{1}{2} (a + b)$ .
Three points $A$ , $B$ , $C$ are collinear if $A C = k A B$ for some scalar $k$ , since the two vectors are then parallel and share the point $A$ .

Method

Compute $A B = b - a$ and $A C = c - a$ .
Test whether one is a scalar multiple of the other, stating the value of $k$ explicitly.
Conclude that the two vectors are parallel and share the common point $A$ , so $A$ , $B$ and $C$ are collinear.

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

$A B = 231, A C = 693$
$A C = 3 A B$ parallel, with $k = 3$
The two vectors are parallel and share the point $A$ , so $A$ , $B$ and $C$ lie on one line.

Notes

Parallel on its own is not enough: two parallel vectors with no shared point describe two distinct parallel lines. The concluding sentence must name the common point.
The value of $k$ also gives the ratio in which the points divide the line, so it is worth recording rather than discarding.
A negative $k$ still proves collinearity (the points are parallel, just on opposite sides of $A$ ) — don't reject it as a failure, but do state whether $C$ lies beyond $B$ or on the far side of $A$ .

Dividing a line in a given ratio 1.10

Key results

A point $P$ dividing the line segment $A B$ in the ratio $m : n$ measured from $A$ has position vector $p = a + \frac{m}{m + n} (b - a)$ — the starting point plus the appropriate fraction of the displacement toward the end point.
Expanding gives the symmetric form $p = \frac{n a + m b}{m + n}$ , which is often quicker to evaluate.
The midpoint is the special case $m : n = 1 : 1$ , giving $p = \frac{1}{2} (a + b)$ .

Method

Identify which point the ratio is measured from — this fixes which of $m$ and $n$ is which.
Compute the displacement $A B = b - a$ .
Add the fraction $\frac{m}{m + n}$ of that displacement to $a$ , working component by component.

In practice $A$ is $(2, - 1)$ and $B$ is $(8, 5)$ . Find the point $P$ dividing $A B$ in the ratio $1 : 2$ .

$A B = (66)$
$p = a + \frac{1}{3} A B = (2 - 1) + (22) = (41)$ from $A$ : $\frac{1}{1 + 2}$ of the way

Notes

A ratio of $2 : 1$ from $A$ is the same point as a ratio of $1 : 2$ from $B$ — always state, and check, which end the ratio starts from.
A quick sanity check: the point should be nearer whichever end has the smaller share of the ratio, so $2 : 1$ from $A$ lands two-thirds of the way along, close to $B$ .
Both formulas give exactly the same point — use the symmetric form $\frac{n a + m b}{m + n}$ for a quick numeric answer, and the first form when showing the geometric reasoning is what's required.

Angles, and vectors in context 1.10

Key results

The lengths of all three sides of a triangle can be found as magnitudes of vectors, after which the cosine rule $cos A = \frac{b ^{2} + c ^{2} - a ^{2}}{2 b c}$ gives any angle of the triangle.
In two dimensions the angle a vector makes with a coordinate axis comes directly from its components by right-angled trigonometry.
A particle starting at position vector $r_{0}$ and moving with constant velocity $v$ has position vector $r = r_{0} + t v$ at time $t$ , and speed $∣ v ∣$ .
Forces combine by vector addition: the resultant of several forces is their vector sum, and a particle is in equilibrium exactly when that resultant is the zero vector.

Notes

The scalar (dot) product is not part of OCR H240 A-level Mathematics, so angles between vectors are obtained here from the cosine rule or from right-angled trigonometry rather than from a dot-product formula.
Round angles to 3 significant figures unless the question specifies otherwise, and keep full accuracy in the side lengths until the final step.
For an equilibrium question, resolving the forces into components along two convenient perpendicular directions and setting each component's sum to zero is usually faster than working with the vectors as a whole.

Worked examples

Worked example

Points $A (1, 2, 3)$ and $B (4, 6, 3)$ are given.

Find $A B$ , its magnitude, the midpoint of $A B$ , and the position vector of the point $P$ that divides $A B$ in the ratio $2 : 1$ from $A$ .

Show worked solution

A B = b - a = (4 - 1, 6 - 2, 3 - 3) = (3, 4, 0)

Its magnitude is:

∣ A B ∣ = 3^{2} + 4^{2} + 0^{2} = 9 + 16 = 25 = 5

The midpoint has position vector:

\frac{1}{2} (a + b) = \frac{1}{2} (5, 8, 6) = (\frac{5}{2}, 4, 3)

For $P$ dividing $A B$ in the ratio $2 : 1$ from $A$ :

p = a + \frac{2}{3} (b - a) = (1, 2, 3) + \frac{2}{3} (3, 4, 0) = (3, \frac{14}{3}, 3)

As a check,

∣ A P ∣ = \frac{2}{3} (3, 4, 0) = \frac{2}{3} (5) = \frac{10}{3}

which is indeed two-thirds of $∣ A B ∣ = 5$ .

Worked example

Show that the points $A (2, - 1, 3)$ , $B (5, 1, 7)$ and $C (11, 5, 15)$ are collinear, and find the ratio $A B : BC$ .

Show worked solution

A B = b - a = (3, 2, 4)

and:

A C = c - a = (9, 6, 12)

Since:

(9, 6, 12) = 3 (3, 2, 4)

we have $A C = 3 A B$ , so $A C$ and $A B$ are parallel; as they also share the point $A$ , the three points are collinear.

For the ratio,

BC = c - b = (6, 4, 8) = 2 A B

so $A B : BC = 1 : 2$ .

Confirming with magnitudes:

∣ A B ∣ = 9 + 4 + 16 = 29

and:

∣ BC ∣ = 36 + 16 + 64 = 116 = 229

a ratio of $1 : 2$ as required.

Worked example

The points $A (1, 2, 3)$ , $B (4, 6, 3)$ and $C (1, 6, 6)$ form a triangle.

Find the size of angle $B A C$ , correct to 3 significant figures.

Show worked solution

Find the three side lengths as magnitudes.

A B = (3, 4, 0)

so:

A B = 9 + 16 + 0 = 5

A C = (0, 4, 3)

so:

A C = 0 + 16 + 9 = 5

BC = c - b = (- 3, 0, 3)

so:

BC = 9 + 0 + 9 = 18 = 32

Applying the cosine rule at $A$ , where the side opposite $A$ is $BC$ :

cos (B A C) = \frac{A B ^{2} + A C ^{2} - B C ^{2}}{2 ( A B ) ( A C )} = \frac{25 + 25 - 18}{2 ( 5 ) ( 5 )} = \frac{32}{50} = 0.64

So angle:

B A C = cos^{- 1} (0.64) = 50. 2^{\circ}

(3 s.f.).

The triangle is isosceles with $A B = A C$ , so the other two angles are equal, each:

\frac{18 0 ^{\circ} - 50. 2 ^{\circ}}{2} = 64. 9^{\circ}

Worked example

A particle starts at the point with position vector:

(2 - 1)

and moves with constant velocity:

(34)

m s

$^{- 1}$ .

Find its position after $5$ s, its distance from the origin at that moment, and the time at which it is $25$ m from its starting point.

Show worked solution

The position vector at time $t$ is:

r = (2 - 1) + t (34)

At $t = 5$ :

r = (2 + 15 - 1 + 20) = (1719)

so the particle is at $(17, 19)$ .

Its distance from the origin is:

1 7^{2} + 1 9^{2} = 289 + 361 = 650 = 25.5

(3 s.f.).

The displacement from the start is:

t (34)

whose magnitude is:

t 3^{2} + 4^{2} = 5 t

— the speed is $5$ m s $^{- 1}$ — so $5 t = 25$ gives $t = 5$ s.

Note that this is not the same as the distance from the origin, since the particle did not start there.