J4 - Position vectors and geometry | AQA A-Level Mathematics

Use position vectors to locate points, subtract them to find a displacement, and measure the distance between fixed or moving points. You will also locate midpoints and ratio points, while keeping the difference between a whole line and a segment clear.

Position and displacement vectors

If a map moves its origin, do you need to change the journey from your house to school? The coordinates change, but the directed journey does not. Position and displacement answer two different questions: “Where is the point?” and “What change takes us between the points?”

You already know

A vector is a signed change along perpendicular axes. Add vectors head-to-tail or componentwise, and use the square root of the sum of squared components for a length.

An origin lets us use one of those directed arrows to locate a point itself.

Position vector

The position vector of $A$ relative to an origin $O$ is $\overrightarrow{OA}$ . Its components are the coordinates of $A$ in the chosen axes. The displacement $\overrightarrow{AB}$ records the directed change from $A$ to $B$ .

For a small concrete example, positions $\mathbf a=\mathbf i+2\mathbf j$ and $\mathbf b=5\mathbf i+4\mathbf j$ are joined by the displacement $4\mathbf i+2\mathbf j$ . Add that change to the starting position and you reach the ending position.

Position vectors OA and OB, with displacement AB=b-a on equal-scale axes

The arrow relationship is $\overrightarrow{OA}+\overrightarrow{AB}=\overrightarrow{OB}$ . Rearranging gives a subtraction rule; it does not say that their three scalar lengths add.

End minus start

If $\overrightarrow{OA}=\mathbf a$ and $\overrightarrow{OB}=\mathbf b$ , then

\overrightarrow{AB}=\mathbf b-\mathbf a\qquad \overrightarrow{BA}=\mathbf a-\mathbf b

Not in the formula booklet: learn it.

If the new origin $O'$ has old position vector $\mathbf c$ , while the axes keep their directions and unit lengths, the new positions are $\mathbf a-\mathbf c$ and $\mathbf b-\mathbf c$ . Their difference is

(\mathbf b-\mathbf c)-(\mathbf a-\mathbf c)=\mathbf b-\mathbf a

The origin shift cancels. You could move the origin a million units away and the displacement between the two points would remain exactly the same. This follows algebraically for any $\mathbf c$ , rather than just for one chosen map.

Worked example 1: a directed change in space

Question. Relative to the same origin, $A$ and $B$ have positions $\mathbf a=\begin{bmatrix}-1\\2\\4\end{bmatrix}$ and $\mathbf b=\begin{bmatrix}3\\-1\\6\end{bmatrix}$ . Find $\overrightarrow{AB}$ .

Plan. Subtract the starting position $\mathbf a$ from the ending position $\mathbf b$ .

\begin{aligned} \overrightarrow{AB}&=\mathbf b-\mathbf a && \text{end minus start}\\ &=\begin{bmatrix}3-(-1)\\-1-2\\6-4\end{bmatrix} && \text{corresponding coordinates}\\ &=\begin{bmatrix}4\\-3\\2\end{bmatrix} \end{aligned}

Why this step? The displacement must be the vector that takes you from $A$ to $B$ , rather than from the origin to either point.

Answer. $\overrightarrow{AB}=4\mathbf i-3\mathbf j+2\mathbf k$ .

Check. $(-1,2,4)+(4,-3,2)=(3,-1,6)$ , the position of $B$ .

Change only the direction of travel in the next example. Neither point moves, but the joining arrow reverses.

Worked example 2: the return displacement

Question. For $A=(-1,2,4)$ and $B=(3,-1,6)$ , find $\overrightarrow{BA}$ .

Plan. Make $B$ the start and $A$ the end before subtracting.

\begin{aligned} \overrightarrow{BA}&=\mathbf a-\mathbf b && \text{new end minus new start}\\ &=\begin{bmatrix}-1-3\\2-(-1)\\4-6\end{bmatrix}\\ &=\begin{bmatrix}-4\\3\\-2\end{bmatrix} \end{aligned}

Why this step? Reversing travel changes every signed component, even though it does not change the distance between the points.

Answer. $\overrightarrow{BA}=-4\mathbf i+3\mathbf j-2\mathbf k=-\overrightarrow{AB}$ .

Check. Adding this to $(3,-1,6)$ returns $(-1,2,4)$ .

You can always check a displacement by adding it to the stated starting position.

Watch out

Wrong: $\overrightarrow{AB}=\mathbf a+\mathbf b$ . Two positions both start at the origin, so adding them does not generally join their endpoints. Right: $\mathbf a+\overrightarrow{AB}=\mathbf b$ , hence $\overrightarrow{AB}=\mathbf b-\mathbf a$ .

Use the same end-minus-start test in the plane.

Distance between fixed or moving points

The displacement tells you the direction between two points; its magnitude gives their distance apart. For $A=(-1,2,4)$ and $B=(3,-1,6)$ , the joining vector is $(4,-3,2)$ , so the distance is $\sqrt{16+9+4}=\sqrt{29}$ .

Distance from position vectors

For positions $\mathbf a=a_1\mathbf i+a_2\mathbf j+a_3\mathbf k$ and $\mathbf b=b_1\mathbf i+b_2\mathbf j+b_3\mathbf k$ ,

AB=|\mathbf b-\mathbf a|=\sqrt{(b_1-a_1)^2+(b_2-a_2)^2+(b_3-a_3)^2}

Omit the third term in two dimensions. Not in the formula booklet: learn it.

The scalar $AB$ is a length; $\overrightarrow{AB}$ is a vector. Reversing the displacement changes every component's sign, but squaring removes those signs, so $|\mathbf b-\mathbf a|=|\mathbf a-\mathbf b|$ . Opposite corners of a cube of side $s$ are therefore $\sqrt{s^2+s^2+s^2}=s\sqrt3$ apart: the same three-dimensional rule in a familiar solid.

When positions depend on time, evaluate both at the same time. Comparing one object's position now with another object's position later does not measure their simultaneous separation.

Worked example 3: simultaneous separation

Question. Two moving points have positions $\mathbf r_A(t)=(t+1)\mathbf i+2t\mathbf j$ and $\mathbf r_B(t)=5\mathbf i+(t+2)\mathbf j$ , in metres, where $t\ge0$ is in seconds. Find their separation at $t=3$ .

Plan. Evaluate both positions at $t=3$ , subtract, then take the magnitude.

\begin{aligned} \mathbf r_A(3)&=4\mathbf i+6\mathbf j && \text{first simultaneous position}\\ \mathbf r_B(3)&=5\mathbf i+5\mathbf j && \text{second simultaneous position}\\ \mathbf r_B(3)-\mathbf r_A(3)&=\mathbf i-\mathbf j && \text{joining vector}\\ AB&=\sqrt{1^2+(-1)^2}=\sqrt2\,\mathrm m && \text{its length} \end{aligned}

Why this step? A distance between moving objects needs their two locations at one common instant.

Answer. Their separation is $\sqrt2\,\mathrm m$ .

Check. From $(4,6)$ to $(5,5)$ , each coordinate changes by one unit in magnitude, matching a unit-square diagonal.

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J4 - Position vectors and geometry

Position and displacement vectors

Distance between fixed or moving points

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