Mv4–Mv6 - Position vectors, distances and vector problems | OCR A-Level Mathematics B (MEI)

Connect Cartesian coordinates with position vectors, calculate distances and locate points in a given ratio. Build geometric arguments from general vectors and use the same addition rule to find and balance force resultants.

From positions to displacements

A survey team marks a collection point $P$ on a straight path from $A=(-2,1)$ to $B=(4,9)$ , with $AP:PB=3:2$ . Where is $P$ ? A different map origin must not change its physical location, so our method needs to separate positions from movements.

You already know

Vector addition combines component changes; multiplication by a scalar changes each component. The modulus $|(x,y)|=\sqrt{x^2+y^2}$ is a scalar length. Parallel non-zero vectors are scalar multiples.

Choose an origin $O$ . A position vector $\overrightarrow{OA}=\mathbf a$ locates $A$ ; its Cartesian components are the coordinates of $A$ . The displacement $\overrightarrow{AB}$ starts at $A$ , so it is a different arrow from the position vector of $B$ .

Position vectors and displacement from A to B
For the separate example $A=(1,-2)$ , $B=(7,6)$ , the position arrows begin at $O$ , while $\overrightarrow{AB}$ has components $(6,8)$ and length $10$ .

Travelling $A\to O\to B$ means reversing $\mathbf a$ , then adding $\mathbf b$ . That route supplies the subtraction rule.

From positions to a displacement

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

\overrightarrow{AB}=\mathbf b-\mathbf a,\qquad AB=|\mathbf b-\mathbf a|

For Cartesian points, $AB=\sqrt{(b_1-a_1)^2+(b_2-a_2)^2}$ . The first expression is a vector; the second is its scalar length.

Not in the formula sheet: learn it.

There is a neat cancellation hidden here. If a new origin has old position $\mathbf c$ , the new positions are $\mathbf a-\mathbf c$ and $\mathbf b-\mathbf c$ , but their difference is still $\mathbf b-\mathbf a$ . You can change every coordinate on the map without changing any journey between fixed points.

Worked example 1: the survey displacement

Question. Find $\overrightarrow{AB}$ and $AB$ for $A=(-2,1)$ , $B=(4,9)$ .

Plan. Subtract the coordinates of $A$ from those of $B$ , then find the modulus.

\begin{aligned} \overrightarrow{AB}&=(4,9)-(-2,1)=(6,8) &&\text{finish minus start}\\ AB&=\sqrt{6^2+8^2}=10 &&\text{length of the displacement} \end{aligned}

Why this step? A distance between points depends on their coordinate differences, not their separate distances from $O$ .

Answer. $\overrightarrow{AB}=6\mathbf i+8\mathbf j$ , $AB=10$ .

Check. $(-2,1)+(6,8)=(4,9)$ , the required finish.

Change only the origin. Put the new origin at the point whose old coordinates are $(3,-2)$ .

Worked example 2: the same journey on a different map

Question. Find the new positions of $A$ and $B$ , then find their displacement and distance in the new coordinates.

Plan. Subtract the new origin's old position from both points, then subtract the new endpoint positions.

\begin{aligned} \mathbf a'&=(-2,1)-(3,-2)=(-5,3) &&\text{new origin to }A\\ \mathbf b'&=(4,9)-(3,-2)=(1,11) &&\text{new origin to }B\\ \overrightarrow{AB}&=(1,11)-(-5,3)=(6,8) &&\text{common shift cancels}\\ AB&=\sqrt{36+64}=10 &&\text{same physical distance} \end{aligned}

Why this step? Moving the origin changes positions, not the vector between the two fixed points.

Answer. New positions $(-5,3)$ , $(1,11)$ ; displacement $(6,8)$ , distance $10$ .

Check. Adding $(3,-2)$ back to each new position recovers the old coordinates.

The displacement reverses sign for the return trip, but its modulus remains $10$ . Neither changing the origin nor reversing a route makes a distance negative.

Watch out

Wrong: find $AB$ by subtracting $|\mathbf a|$ from $|\mathbf b|$ . Right: subtract the vectors first, then take the modulus. The two endpoint lengths refer to $O$ , not to the connecting segment.

Try another common change of origin, keeping the physical points fixed.

Locating points along a segment

A ratio divides the whole displacement into equal shares. Once you know the movement from $A$ , include $A$ 's position to locate the endpoint of that movement.

Internal division of a segment

If $AP:PB=m:n$ , with $m,n>0$ and $P$ between $A$ and $B$ , then

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

More generally, a fraction $t$ from $A$ gives $\mathbf p=(1-t)\mathbf a+t\mathbf b$ . The midpoint uses $t=1/2$ . Values $t<0$ or $t>1$ go beyond the segment.

Not in the formula sheet: learn it.

The coefficient of the movement is $m/(m+n)$ , not $m/n$ . In the diagram's example $A=(1,-2)$ , $B=(7,6)$ , ratio $2:1$ gives $(1,-2)+\frac23(6,8)=(5,10/3)$ ; the two displacement pieces are in the required ratio.

Worked example 3: the collection point

Question. For $A=(-2,1)$ , $B=(4,9)$ , find $P$ with $AP:PB=3:2$ internally.

Plan. Move three of the five equal shares of $\overrightarrow{AB}$ from $A$ .

\begin{aligned} t&=\frac{3}{3+2}=\frac35 &&\text{fraction of the whole}\\ \mathbf p&=(-2,1)+\frac35(6,8) &&\text{position plus displacement}\\ &=\left(\frac85,\frac{29}{5}\right) &&\text{combine exact fractions} \end{aligned}

Why this step? The first ratio number measures the part from $A$ , so the fraction must also be measured from $A$ .

Answer. $P=(8/5,29/5)$ .

Check. $\overrightarrow{AP}=(18/5,24/5)$ and $\overrightarrow{PB}=(12/5,16/5)$ , so the first is $3/2$ times the second.

Change only the ratio from $3:2$ to $2:3$ . The point moves to the other side of the midpoint.

Worked example 4: exchanging the two shares

Question. For the same endpoints, find $P$ with $AP:PB=2:3$ internally.

Plan. Add two-fifths of the displacement to $A$ .

\begin{aligned} \mathbf p&=(-2,1)+\frac25(6,8) &&\text{two of five shares}\\ &=\left(\frac25,\frac{21}{5}\right) &&\text{add coordinates} \end{aligned}

Why this step? The whole still has five shares; only the number travelled from $A$ changes.

Answer. $P=(2/5,21/5)$ .

Check. $\overrightarrow{AP}=(12/5,16/5)$ , $\overrightarrow{PB}=(18/5,24/5)$ , giving ratio $2:3$ .

This calculation uses a specified fraction to locate a point. There is no need to introduce vector equations of lines. When the endpoints are letters, exactly the same weighted combination works.

Watch out

Wrong: report only $\frac25(\mathbf b-\mathbf a)$ as the position of $P$ . That is $\overrightarrow{AP}$ . Right: add $\mathbf a$ so the position vector starts at the origin.

Apply the same fraction to a new segment.

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Mv4–Mv6 - Position vectors, distances and vector problems

From positions to displacements

Locating points along a segment

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