Vectors in two and three dimensionsEdexcel International A Level Maths: Revision notes

Section 1

Vectors and components

A vector has magnitude and direction. In three dimensions, $\mathbf{i}$ , $\mathbf{j}$ , $\mathbf{k}$ are unit vectors along the $x$ , $y$ , $z$ axes, so $\mathbf{a}=a_1\mathbf{i}+a_2\mathbf{j}+a_3\mathbf{k}$ , also written as a column vector. In two dimensions there are only $\mathbf{i}$ and $\mathbf{j}$ . Add and subtract component by component: $(2\mathbf{i}-\mathbf{j}+3\mathbf{k})+(\mathbf{i}+4\mathbf{j})=3\mathbf{i}+3\mathbf{j}+3\mathbf{k}$ . Multiplying by a scalar $\lambda$ multiplies every component, giving a vector parallel to the original: $\mathbf{b}=\lambda\mathbf{a}$ means $\mathbf{a}$ and $\mathbf{b}$ are parallel, and $\lambda<0$ reverses direction.

Key termsvectorscalarparallel vectors

Common mistake

Treating $\mathbf{i}$ , $\mathbf{j}$ , $\mathbf{k}$ components as if they could be combined across directions: $2\mathbf{i}+3\mathbf{j}$ cannot be simplified.

Section 2

Magnitude and unit vectors

The magnitude of $\mathbf{a}=a_1\mathbf{i}+a_2\mathbf{j}+a_3\mathbf{k}$ is $|\mathbf{a}|=\sqrt{a_1^2+a_2^2+a_3^2}$ , from Pythagoras applied twice. For $2\mathbf{i}+3\mathbf{j}-6\mathbf{k}$ , $|\mathbf{a}|=\sqrt{4+9+36}=7$ . A unit vector has magnitude 1. The unit vector in the direction of $\mathbf{a}$ is $\hat{\mathbf{a}}=\frac{\mathbf{a}}{|\mathbf{a}|}$ . For $\overrightarrow{AB}=3\mathbf{i}+4\mathbf{j}-4\mathbf{k}$ , $\hat{\mathbf{a}}=\frac{1}{\sqrt{41}}(3\mathbf{i}+4\mathbf{j}-4\mathbf{k})$ . A vector of length $k$ in that direction is $k\hat{\mathbf{a}}$ .

Key termsmagnitudeunit vector

Common mistake

Adding the components instead of squaring them: $|2\mathbf{i}+3\mathbf{j}-6\mathbf{k}|=7$ , not $2+3-6$ .

Exam tip

Check a unit vector: its components squared should sum to 1.

Section 3

Position vectors and AB = b − a

The position vector of a point $P$ relative to the origin $O$ is $\overrightarrow{OP}=\mathbf{p}$ , with components equal to the coordinates of $P$ . By the triangle law, $\overrightarrow{OA}+\overrightarrow{AB}=\overrightarrow{OB}$ , so $\overrightarrow{AB}=\mathbf{b}-\mathbf{a}.$ The midpoint of $AB$ has position vector $\frac12(\mathbf{a}+\mathbf{b})$ . To go beyond $B$ so that $B$ is the midpoint of $AC$ , use $\mathbf{c}=2\mathbf{b}-\mathbf{a}$ . Equal vectors have the same length and direction, so $\overrightarrow{BD}=\overrightarrow{OA}$ shows $BD$ is parallel and equal to $OA$ , the test for a parallelogram.

Key termsposition vectortriangle lawmidpoint

Common mistake

Writing $\overrightarrow{AB}=\mathbf{a}-\mathbf{b}$ . It is destination minus start: $\mathbf{b}-\mathbf{a}$ .

Section 4

Distance between two points

The distance between $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ is the magnitude of $\overrightarrow{AB}$ : $d^2=(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2.$ For $A(-1,2,3)$ and $B(3,-2,5)$ : $d^2=16+16+4=36$ , so $d=6$ . Differences are squared, so the order of subtraction does not matter, but the signs inside each bracket do. Use this to compare lengths: if $AB=AC=\sqrt{41}$ the triangle is isosceles, and the median from $A$ to the midpoint of $BC$ is perpendicular to $BC$ .

Key termsdistance

Common mistake

Subtracting the squared terms or not taking the square root at the end.

Section 5

Using vectors in geometry

Combine the ideas: find $\overrightarrow{OB}=\overrightarrow{OA}+\overrightarrow{OC}$ in a parallelogram $OABC$ ; find midpoints by averaging; find lengths using magnitude; show parallelism by writing one vector as a multiple of another; and justify shape properties using your results in a sentence. Example: $OABC$ with $\overrightarrow{OA}=4\mathbf{i}+\mathbf{j}+2\mathbf{k}$ , $\overrightarrow{OC}=-\mathbf{i}+3\mathbf{j}+\mathbf{k}$ gives $\overrightarrow{OB}=3\mathbf{i}+4\mathbf{j}+3\mathbf{k}$ , and the midpoint of $AB$ is $\frac72\mathbf{i}+\frac52\mathbf{j}+\frac52\mathbf{k}$ with $OM=\frac{3\sqrt{11}}{2}$ .

Exam tip

In a 'show that' question, finish with a sentence linking your vectors to the geometric statement.

That's the notes covered.

Carry on to the next subtopic.

Exam questions on Vectors in two and three dimensions

The points $A$ and $B$ have position vectors $\mathbf{a}=2\mathbf{i}-\mathbf{j}+3\mathbf{k}$ and $\mathbf{b}=5\mathbf{i}+3\mathbf{j}-\mathbf{k}$ relative to the origin $O$ .
Find a unit vector in the direction of $\overrightarrow{AB}$ .2 marks
The points $A$ and $B$ have coordinates $(-1,2,3)$ and $(3,-2,5)$ relative to the origin $O$ , and $M$ is the midpoint of $AB$ .
The point $C$ is such that $B$ is the midpoint of $AC$ . Find the position vector of $C$ .2 marks
Triangle $ABC$ has vertices $A(1,2,-1)$ , $B(4,-2,3)$ and $C(5,5,3)$ , where the coordinates are in metres. A calculator may be used.
Show that triangle $ABC$ is isosceles.3 marks

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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).