Position vectors and geometry · Vectors

The position vector of a point $A$ is $\mathbf a = \overrightarrow{OA}$ , the vector from the origin to $A$ . The vector from one point to another is end minus start (see vectors):

From A to BAB = b − a: end minus start

With that one fact, many geometry questions become simple arithmetic.

The fourth vertex of a parallelogram

In a parallelogram $ABCD$ , opposite sides are equal and parallel: $\overrightarrow{AB} = \overrightarrow{DC}$ . So $\mathbf b - \mathbf a = \mathbf c - \mathbf d$ , which gives $\mathbf d = \mathbf a + \mathbf c - \mathbf b$ :

The fourth vertexd = a + c − b: go round the shape in order

Worked example · WAEC 2019

WAEC 2019 · Paper 2 · Q8

$A(-3, 1)$ , $B(1, 2)$ , $C(0, -1)$ and $D(x, y)$ are the vertices of a parallelogram $ABCD$ . Using the vector method, determine the coordinates of $D$ .

Opposite sides
- ${\overrightarrow{AB} = \overrightarrow{DC}}$ , so ${\mathbf b - \mathbf a = \mathbf c - \mathbf d}$ .
- So ${\mathbf d = \mathbf a + \mathbf c - \mathbf b}$ .
Think first. In ABCD, which side equals AB?
Substitute
- ${\mathbf d = \begin{pmatrix} -3 \\ 1 \end{pmatrix} + \begin{pmatrix} 0 \\ -1 \end{pmatrix} - \begin{pmatrix} 1 \\ 2 \end{pmatrix} = \begin{pmatrix} -4 \\ -2 \end{pmatrix}}$ .
- So $D$ is ${(-4, -2)}$ .

More: parallelograms

Dividing a line in a ratio

If $M$ is on $AB$ with $AM : MB = m : n$ , then $M$ is $\frac{m}{m + n}$ of the way from $A$ to $B$ :

Dividing AB in the ratio 2 : 1OM = (n a + m b) ÷ (m + n)

\overrightarrow{OM} = \mathbf a + \frac{m}{m + n}(\mathbf b - \mathbf a) = \frac{n\mathbf a + m\mathbf b}{m + n}

Notice the cross-over: $\mathbf a$ is multiplied by $n$ , the part next to $B$ .

Dividing a line in a ratioSet m and n

3i + 5/2 jOM3/4fraction of the way from A

m = 3n = 1

AM : MB = 3 : 1, so M is 3/4 of the way from A to B. OM = (a + 3b) ÷ 4 = ((−3i − 2j) + 3(5i + 4j)) ÷ 4 = 3i + 5/2 j. The number next to a is n, the part of the line nearer B.

Worked example · WAEC 2018

WAEC 2018 · Paper 2 · Q7

Two points $X$ and $Y$ have position vectors $\mathbf x = 2\mathbf i - 3\mathbf j$ and $\mathbf y = -\mathbf i + 2\mathbf j$ . Find the position vector of the point $M$ on $\overline{XY}$ such that $|\overrightarrow{XM}| : |\overrightarrow{MY}| = 3 : 2$ .

The ratio
- $M$ is ${\frac35}$ of the way from $X$ to $Y$ .
Think first. XM : MY = 3 : 2. How far from X to Y is M?
The position vector
- ${\overrightarrow{OM} = \frac{2\mathbf x + 3\mathbf y}{5}}$ .
- ${= \frac{2(2\mathbf i - 3\mathbf j) + 3(-\mathbf i + 2\mathbf j)}{5}}$ .
- ${= \frac{(4 - 3)\mathbf i + (-6 + 6)\mathbf j}{5} = \frac15\mathbf i}$ .

More: lines and their intersections

Collinear points

Three points $P$ , $Q$ and $R$ lie on a straight line if $\overrightarrow{PQ}$ and $\overrightarrow{QR}$ are parallel (one is a multiple of the other), because they share the point $Q$ .

Worked example · WAEC 2011

WAEC 2011 · Paper 2 · Q16 (a)

The position vectors of points $P$ , $Q$ and $R$ are $5\mathbf i + 3\mathbf j$ , $8\mathbf i - \mathbf j$ and $11\mathbf i - 5\mathbf j$ respectively. (i) Show that $P$ , $Q$ and $R$ are collinear. (ii) Find the scalars $k_1$ and $k_2$ such that $37\mathbf i - \mathbf j = k_1\mathbf p + k_2\mathbf r$ where $\mathbf p$ and $\mathbf r$ are the position vectors of $P$ and $R$ respectively.

Collinear
- ${\overrightarrow{PQ} = (8 - 5)\mathbf i + (-1 - 3)\mathbf j = 3\mathbf i - 4\mathbf j}$ .
- ${\overrightarrow{QR} = (11 - 8)\mathbf i + (-5 + 1)\mathbf j = 3\mathbf i - 4\mathbf j}$ .
- They are equal, so parallel, and they share $Q$ : $P$ , $Q$ and $R$ are collinear.
Think first. Find PQ and QR.
The scalars
- The $\mathbf i$ parts: ${5k_1 + 11k_2 = 37}$ . The $\mathbf j$ parts: ${3k_1 - 5k_2 = -1}$ .
- From the second: ${k_1 = \frac{5k_2 - 1}{3}}$ .
- Substitute: ${\frac{5(5k_2 - 1)}{3} + 11k_2 = 37}$ , so ${58k_2 = 116}$ and ${k_2 = 2}$ .
- Then ${k_1 = \frac{10 - 1}{3} = 3}$ .
Think first. Match the i parts and the j parts of k₁p + k₂r.

More: collinear points

Your turn

WAEC 2023 · Paper 2 · Q15 (a)

Vectors

The position vectors of points $P$ , $Q$ and $R$ with respect to the origin are $(-5\mathbf i + \mathbf j)$ , $(2\mathbf i - 4\mathbf j)$ and $(4\mathbf i + 3\mathbf j)$ respectively. If $PQRS$ is a parallelogram, find the:

(a)
position vector of $S$ ;
Show the answer
$-3\mathbf i + 8\mathbf j$

Try it on a graph

Plot the curves, move them, and read values off the graph.

Worked solution (try it first)

(a)

In $PQRS$ , $\overrightarrow{PS} = \overrightarrow{QR}$
$= (4 - 2)\mathbf i + (3 + 4)\mathbf j$
$= 2\mathbf i + 7\mathbf j$ .
So $\mathbf s = \mathbf p + \overrightarrow{PS}$
$= (-5 + 2)\mathbf i + (1 + 7)\mathbf j$
$= -3\mathbf i + 8\mathbf j$ .

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