10 Vectors — revision question pack

5 specification points · notes, questions, answers and worked methods

Checked against Edexcel 9MA0 section 10. Review basis: the qualification registry sourced from the Pearson Edexcel Level 3 Advanced GCE in Mathematics (9MA0) specification; registry verification recorded 11 July 2026.

How this checking works

10.1 · Use vectors in two dimensions and in three dimensions.

Explanation

A vector records magnitude and direction; in two or three dimensions it can be written in component form or with the unit vectors $\mathbf{i}$ , $\mathbf{j}$ and $\mathbf{k}$ .
Work component by component, keeping the $x$ , $y$ and $z$ entries aligned; subtract the initial point from the final point to form a displacement vector.
For example, from $A(1,2,3)$ to $B(4,0,5)$ the displacement is $\overrightarrow{AB}=(3,-2,2)$ .
A vector has no fixed location, whereas a point does; a common error is to confuse the coordinates of an endpoint with the components of the displacement leading to it.

Worked example

Given $\mathbf{a}=(2,-1,3)$ and $\mathbf{b}=(-1,4,2)$ , find $2\mathbf{a}-\mathbf{b}$ .

1.First $2\mathbf{a}=(4,-2,6)$ .
2.Subtract corresponding components of $\mathbf{b}$ to get $(4-(-1),-2-4,6-2)=(5,-6,4)$ .

Answer: $2\mathbf{a}-\mathbf{b}=(5,-6,4)$

Common mistakes

Don't change the coordinate order between vectors, so components from different axes are combined.
Don't combine vector components inconsistently, applying a scalar to only one coordinate.

Exam tip

Perform each vector operation component by component and preserve the coordinate order throughout.

Tier 1 · Easy

1.
The point $A$ has coordinates $(-2,5)$ and the point $B$ has coordinates $(4,1)$ . Find $\overrightarrow{AB}$ .
(2)
(Total for Question 1 is 2 marks)
2.
Given $\mathbf{a}=-4\mathbf{i}+7\mathbf{j}-2\mathbf{k}$ , write $\mathbf{a}$ in component form and find $3\mathbf{a}$ .
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1.
Given $\mathbf{a}=(1,2,-1)$ , $\mathbf{b}=(2,-1,3)$ and $\mathbf{c}=(5,0,5)$ , find scalars $\lambda$ and $\mu$ such that $\lambda\mathbf{a}+\mu\mathbf{b}=\mathbf{c}$ .
(4)
(Total for Question 1 is 4 marks)
2.
Point $A$ has coordinates $(2,-1,4)$ and $\overrightarrow{AB}=(-3,5,2)$ . Find the coordinates of $B$ and find $\overrightarrow{BA}$ . Verify that $\overrightarrow{AB}+\overrightarrow{BA}=\mathbf{0}$ .
(3)
(Total for Question 2 is 3 marks)
3.
Given $\mathbf{a}=(p,2q,-3)$ , $\mathbf{b}=(4,-1,r)$ and $\mathbf{a}+\mathbf{b}=(7,5,2)$ , find $p$ , $q$ and $r$ . Hence find $2\mathbf{a}-\mathbf{b}$ .
(4)
(Total for Question 3 is 4 marks)

Tier 3 · Hard

1.
Let $\mathbf{p}=(1,1,2)$ and $\mathbf{q}=(2,-1,1)$ . By writing $(u,v,w)=\lambda\mathbf{p}+\mu\mathbf{q}$ , show that $(u,v,w)$ lies in the span of $\mathbf{p}$ and $\mathbf{q}$ if and only if $u+v-w=0$ . Hence find the value of $k$ for which $(7,1,k)$ lies in this span and the value of $k$ for which $(5,k,7)$ lies in this span. Can the same value of $k$ satisfy both conditions?
(4)
(Total for Question 1 is 4 marks)
2.
A point starts at $A=(1,-2,5)$ and undergoes, in order, displacements $\mathbf{u}=(3,1,-2)$ , $\lambda\mathbf{v}$ where $\mathbf{v}=(-1,2,1)$ , and $\mathbf{w}=(2,-3,-4)$ . The final point lies in the plane $z=0$ . Find $\lambda$ , the final point and the single displacement from $A$ to the final point.
(5)
(Total for Question 2 is 5 marks)
3.
The vectors $\mathbf{a}$ and $\mathbf{b}$ satisfy $2\mathbf{a}+\mathbf{b}=(7,1,5)$ and $\mathbf{a}-2\mathbf{b}=(1,-12,10)$ . Find $\mathbf{a}$ and $\mathbf{b}$ , and verify both vector equations.
(4)
(Total for Question 3 is 4 marks)
4.
Let $\mathbf{u}=(1,1,0)$ , $\mathbf{v}=(0,1,1)$ and $\mathbf{w}=(1,0,1)$ . A displacement $(7,8,9)$ is formed as $p\mathbf{u}+q\mathbf{v}+r\mathbf{w}$ . Find $p$ , $q$ and $r$ , and verify the displacement component by component.
(5)
(Total for Question 4 is 5 marks)
5.
The vectors $\mathbf{a}=(p,q,r)$ , $\mathbf{b}=(q,r,p)$ and $\mathbf{c}=(r,p,q)$ satisfy $\mathbf{a}+\mathbf{b}+\mathbf{c}=(13,13,13)$ and $\mathbf{a}-\mathbf{b}=(1,-3,2)$ . Find $\mathbf{a}$ , $\mathbf{b}$ and $\mathbf{c}$ , and verify both vector equations.
(5)
(Total for Question 5 is 5 marks)

10.2 · Calculate the magnitude and direction of a vector and convert between component form and magnitude/direction form.

Explanation

For $\mathbf{v}=(a,b)$ , its magnitude is $|\mathbf{v}|=\sqrt{a^2+b^2}$ and a direction angle must be stated relative to a specified axis or bearing convention.
A vector of magnitude $r$ at angle $\theta$ anticlockwise from the positive $x$ -axis has components $(r\cos\theta,r\sin\theta)$ ; use signs or a quadrant-aware angle calculation when reversing the process.
For example, magnitude $10$ at $30^\circ$ above the positive $x$ -axis gives $(10\cos30^\circ,10\sin30^\circ)=(5\sqrt3,5)$ .
The value from $\tan^{-1}(b/a)$ alone can select the wrong quadrant; a common error is to report an acute reference angle without checking the component signs.
A unit vector in the direction of non-zero $\mathbf a$ is $\mathbf a/|\mathbf a|$ ; divide every component by the vector's magnitude.

Worked example

A vector has magnitude $14$ and direction $120^\circ$ anticlockwise from the positive $x$ -axis. Write it in exact component form.

1.Use $(14\cos120^\circ,14\sin120^\circ)$ .
2.Since $\cos120^\circ=-\frac12$ and $\sin120^\circ=\frac{\sqrt3}{2}$ , the vector is $(-7,7\sqrt3)$ .

Answer: $(-7,7\sqrt3)$

Common mistakes

Don't normalise a vector by dividing by the sum of its components rather than by its magnitude.
Don't use sine for the horizontal component and cosine for the vertical component despite the stated reference axis.

Exam tip

Draw the direction from the positive x-axis, then use cosine horizontally and sine vertically with correct signs.

Tier 1 · Easy

1.
Find the magnitude and direction of the vector $(3,4)$ , giving the direction anticlockwise from the positive $x$ -axis to $1$ decimal place.
(3)
(Total for Question 1 is 3 marks)
2.
Find an exact unit vector in the direction of $(-8,6)$ .
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1.
Find the magnitude and direction of the vector $(-6,8)$ , measuring the direction anticlockwise from the positive $x$ -axis and giving the direction to $3$ significant figures.
(3)
(Total for Question 1 is 3 marks)
2.
A displacement has magnitude $12$ and bearing $150^\circ$ , where bearings are measured clockwise from north. Taking east and north as the positive component directions, express the displacement in exact component form.
(3)
(Total for Question 2 is 3 marks)
3.
A vector has magnitude $10$ . Its horizontal component is twice its vertical component, and both components are negative. Find the exact component form and the direction angle anticlockwise from the positive $x$ -axis, giving the angle to $1$ decimal place.
(4)
(Total for Question 3 is 4 marks)

Tier 3 · Hard

1.
A two-dimensional vector has magnitude $13$ , and the cosine of the angle it makes with the positive $x$ -axis is $\frac5{13}$ . Find all possible component forms and explain the ambiguity.
(4)
(Total for Question 1 is 4 marks)
2.
The vector $\mathbf{v}=(k,k-7)$ has magnitude $\sqrt{85}$ . Its direction angle, measured anticlockwise from the positive $x$ -axis, lies between $180^\circ$ and $270^\circ$ . Find $k$ and the direction angle of $\mathbf{v}$ , giving the angle to $1$ decimal place.
(5)
(Total for Question 2 is 5 marks)
3.
Vector $\mathbf{a}$ has magnitude $8$ and direction $30^\circ$ anticlockwise from the positive $x$ -axis. Vector $\mathbf{b}$ has magnitude $6$ and direction $150^\circ$ anticlockwise from the positive $x$ -axis. Find the exact component form of $\mathbf{a}+\mathbf{b}$ , its exact magnitude and its direction angle to $1$ decimal place.
(5)
(Total for Question 3 is 5 marks)
4.
The vector $\mathbf{a}=(3,4)$ . A vector $\mathbf{b}$ has magnitude $5$ , and $\mathbf{a}+\mathbf{b}$ points along the positive $x$ -axis. Find $\mathbf{b}$ and the magnitude of the resultant. Give the direction angle of $\mathbf{b}$ to $1$ decimal place, measured anticlockwise from the positive $x$ -axis.
(5)
(Total for Question 4 is 5 marks)
5.
The vectors $\mathbf{a}=(\sqrt3,1)$ and $\mathbf{b}=(-1,\sqrt3)$ have direction angles measured anticlockwise from the positive $x$ -axis. A vector $\mathbf{v}$ has magnitude $8$ and points along the internal bisector of the smaller angle between the directions of $\mathbf{a}$ and $\mathbf{b}$ . Find the direction angle of $\mathbf{v}$ and its exact component form.
(5)
(Total for Question 5 is 5 marks)

10.3 · Add vectors diagrammatically and perform the algebraic operations of vector addition and multiplication by scalars, and understand their geometrical interpretations.

Explanation

Vector addition combines successive displacements: placing the tail of $\mathbf{b}$ at the head of $\mathbf{a}$ makes the resultant from the first tail to the final head equal to $\mathbf{a}+\mathbf{b}$ .
Add corresponding components and multiply every component by a scalar; subtraction is addition of the opposite vector.
The vector $k\mathbf{a}$ is parallel to $\mathbf{a}$ , has magnitude $|k||\mathbf{a}|$ , and points in the reverse direction when $k<0$ .
A common error is to multiply only one component by a scalar or to draw vectors head-to-head instead of using a head-to-tail or parallelogram construction.

The head-to-tail construction places

\mathbf v

at the head of

2\mathbf u

; the direct arrow is their resultant.

Worked example

Let $\mathbf{u}=(4,1)$ and $\mathbf{v}=(-2,5)$ . Find $2\mathbf{u}+\mathbf{v}$ and describe a head-to-tail construction for this resultant.

1.Calculate $2\mathbf{u}=(8,2)$ , then add $\mathbf{v}$ to obtain $(8-2,2+5)=(6,7)$ .
2.Diagrammatically, translate the vectors without rotating them and place the second $\mathbf{u}$ after the first, followed by $\mathbf{v}$ ; the direct closing vector is the sum.

Answer: $2\mathbf{u}+\mathbf{v}=(6,7)$ .; Place two copies of $\mathbf{u}$ and then one copy of $\mathbf{v}$ head-to-tail; the resultant joins the initial tail to the final head.

Common mistakes

Don't reverse a displacement vector in the diagram but leave its algebraic sign unchanged.
Don't add vector diagrams tail-to-tail without completing the parallelogram or head-to-tail construction.

Exam tip

For a resultant, scale each vector first and join them head-to-tail in the order represented algebraically.

Tier 1 · Easy

1.
Given $\mathbf{u}=(2,-3)$ and $\mathbf{v}=(-5,4)$ , find $\mathbf{u}+\mathbf{v}$ .
(1)
(Total for Question 1 is 1 mark)
2.
Given $\mathbf{u}=(3,-1)$ and $\mathbf{w}=(-6,2)$ , express $\mathbf{w}$ as a scalar multiple of $\mathbf{u}$ and describe its direction and magnitude relative to $\mathbf{u}$ .
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1.
In triangle $ABC$ , $\overrightarrow{AB}=\mathbf{u}$ and $\overrightarrow{AC}=\mathbf{v}$ . The point $M$ is the midpoint of $BC$ . Express $\overrightarrow{AM}$ in terms of $\mathbf{u}$ and $\mathbf{v}$ .
(3)
(Total for Question 1 is 3 marks)
2.
Three successive displacements are $(4,-1)$ , $(-2,5)$ and $(-3,-2)$ . Find their resultant and the single displacement that would return directly to the starting point. Describe how the four vectors form a closed head-to-tail diagram.
(3)
(Total for Question 2 is 3 marks)
3.
Vectors $\overrightarrow{OA}=\mathbf{a}=(5,2)$ and $\overrightarrow{OB}=\mathbf{b}=(1,-3)$ are drawn from the common point $O$ . Find $\mathbf{a}-\mathbf{b}$ and identify the directed segment between $A$ and $B$ that this vector represents.
(3)
(Total for Question 3 is 3 marks)

Tier 3 · Hard

1.
The vectors $\mathbf{a}=(1,2)$ and $\mathbf{b}=(3,-1)$ combine to give $\mathbf{w}=(11,1)$ . Find $\alpha$ and $\beta$ such that $\mathbf{w}=\alpha\mathbf{a}+\beta\mathbf{b}$ , and interpret the result geometrically.
(4)
(Total for Question 1 is 4 marks)
2.
Let $\mathbf{u}=(2,1)$ and $\mathbf{v}=(1,-2)$ . The vector $\mathbf{r}=2\mathbf{u}+\lambda\mathbf{v}$ is parallel to $(1,3)$ and points in the same direction. Find $\lambda$ , express $\mathbf{r}$ as a scalar multiple of $(1,3)$ , and interpret the result geometrically.
(4)
(Total for Question 2 is 4 marks)
3.
In triangle $ABC$ , $\overrightarrow{AB}=\mathbf{a}$ and $\overrightarrow{AC}=\mathbf{b}$ . Point $D$ divides $AB$ internally in the ratio $AD:DB=2:1$ , and point $E$ divides $AC$ internally in the ratio $AE:EC=2:1$ . Express $\overrightarrow{DE}$ in terms of $\mathbf{a}$ and $\mathbf{b}$ . Hence show that $DE$ is parallel to $BC$ and find $DE:BC$ .
(5)
(Total for Question 3 is 5 marks)
4.
In triangle $OAB$ , the position vectors of $A$ and $B$ are $\mathbf{a}$ and $\mathbf{b}$ . Point $M$ is the midpoint of $AB$ , and point $N$ lies on $OM$ with $ON:NM=2:1$ . Point $P$ is the midpoint of $OB$ . Find the position vector of $N$ . Hence prove that $A$ , $N$ and $P$ are collinear and find $AN:NP$ .
(6)
(Total for Question 4 is 6 marks)
5.
The position vectors of the vertices $A$ , $B$ , $C$ and $D$ of a quadrilateral are $\mathbf{a}$ , $\mathbf{b}$ , $\mathbf{c}$ and $\mathbf{d}$ , where $AC$ and $BD$ are not parallel. Points $P$ , $Q$ , $R$ and $S$ are the midpoints of $AB$ , $BC$ , $CD$ and $DA$ , respectively. Use vectors to prove that $PQRS$ is a parallelogram.
(5)
(Total for Question 5 is 5 marks)

10.4 · Understand and use position vectors; calculate the distance between two points represented by position vectors.

Explanation

The position vector of a point $A$ is $\overrightarrow{OA}$ from a fixed origin $O$ ; if these vectors are $\mathbf{a}$ and $\mathbf{b}$ , then $\overrightarrow{AB}=\mathbf{b}-\mathbf{a}$ .
Find a displacement by subtracting position vectors in final-minus-initial order, then find distance by taking the magnitude of that displacement.
A point dividing $AB$ internally in the fraction $t$ from $A$ to $B$ has position vector $\mathbf{a}+t(\mathbf{b}-\mathbf{a})$ .
Distance is a non-negative scalar, not a vector; a common error is to quote $\mathbf{b}-\mathbf{a}$ as the distance without calculating its magnitude.

Worked example

The position vectors of $A$ and $B$ are $(-1,2,4)$ and $(3,-4,7)$ . Calculate the exact distance $AB$ .

1. $\overrightarrow{AB}=(3-(-1),-4-2,7-4)=(4,-6,3)$ .
2.Hence $AB=|\overrightarrow{AB}|=\sqrt{4^2+(-6)^2+3^2}=\sqrt{61}$ .

Answer: $AB=\sqrt{61}$

Common mistakes

Don't square the coordinates of the two points separately instead of squaring their component differences.
Don't find the difference of position vectors but forget to take its Euclidean magnitude.

Exam tip

Subtract the endpoint position vectors in a consistent order, then square and sum every component for the distance.

Tier 1 · Easy

1.
Points $A$ and $B$ have position vectors $(2,-1)$ and $(7,3)$ . Find $\overrightarrow{AB}$ and the exact distance $AB$ .
(3)
(Total for Question 1 is 3 marks)
2.
Point $A$ has position vector $(3,-2,1)$ and $\overrightarrow{AB}=(-5,4,2)$ . Find the position vector of $B$ .
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1.
The position vectors of $A$ and $B$ are $(-2,1,5)$ and $(6,5,-3)$ . Find the midpoint of $AB$ and the point $P$ that divides $AB$ internally in the ratio $AP:PB=3:1$ .
(4)
(Total for Question 1 is 4 marks)
2.
Points $A$ and $B$ have position vectors $(1,2,-1)$ and $(5,-2,3)$ . The point $P$ has position vector $(t,0,1)$ and is equidistant from $A$ and $B$ . Find $t$ and the exact distance $AP$ .
(4)
(Total for Question 2 is 4 marks)
3.
Points $A$ and $B$ have position vectors $(t,2,-1)$ and $(3,-2,5)$ respectively. Given that $AB=2\sqrt{17}$ , find all possible values of $t$ .
(4)
(Total for Question 3 is 4 marks)

Tier 3 · Hard

1.
Points $A$ and $B$ have position vectors $(1,-2,3)$ and $(9,6,-1)$ . The point $M$ divides $AB$ internally in the ratio $AM:MB=3:1$ . Find the position vector of $M$ and the exact distance $OM$ .
(5)
(Total for Question 1 is 5 marks)
2.
The position vectors of $A$ , $B$ and $C$ are $(1,0,2)$ , $(3,1,4)$ and $(-1,2,3)$ respectively. By calculating exact distances, show that triangle $ABC$ is right-angled. Hence find the position vector of the midpoint $M$ of the hypotenuse and verify that $MA=MB=MC$ .
(6)
(Total for Question 2 is 6 marks)
3.
Points $A$ and $B$ have position vectors $(1,-2,3)$ and $(4,4,-3)$ . Point $P$ lies beyond $B$ on the ray from $A$ through $B$ , and $AP:PB=2:1$ . Find the position vector of $P$ and verify the ratio by calculating the exact distances $AP$ and $PB$ .
(5)
(Total for Question 3 is 5 marks)
4.
The position vectors of three points are $\mathbf{a}=(1,2,-1)$ , $\mathbf{b}=(4,6,1)$ and $\mathbf{c}=(10,14,5)$ . Show that their points $A$ , $B$ and $C$ are collinear with $B$ between the other two. Calculate the exact distances $AB$ , $BC$ and $AC$ , and hence find $AB:BC$ .
(5)
(Total for Question 4 is 5 marks)
5.
In three-dimensional coordinate space, let $A=(0,0,0)$ , $B=(4,0,0)$ and $C=(1,k,2)$ . Find every value of $k$ for which triangle $ABC$ is isosceles. State the equal sides in each case.
(6)
(Total for Question 5 is 6 marks)

10.5 · Use vectors to solve problems in pure mathematics and in context (including forces).

Explanation

Vector models turn geometrical displacements or forces into component equations; equilibrium means that the vector sum of all forces is zero.
Choose and state positive coordinate directions, resolve every vector consistently, then equate components or use position-vector relationships.
In a parallelogram with adjacent position vectors $\mathbf{a}$ and $\mathbf{b}$ , the opposite vertex has position vector $\mathbf{a}+\mathbf{b}$ and both diagonals share midpoint $\frac12(\mathbf{a}+\mathbf{b})$ .
A common error is to balance force magnitudes without balancing directions; equal numerical magnitudes do not guarantee equilibrium unless the vector sum is zero.

For the parallelogram with position vectors

\mathbf 0,\mathbf a,\mathbf b,\mathbf a+\mathbf b

, both diagonals share midpoint

(\mathbf a+\mathbf b)/2

Worked example

The points $O,A,B,C$ have position vectors $\mathbf{0},\mathbf{a},\mathbf{b},\mathbf{a}+\mathbf{b}$ respectively. Use vectors to prove that the diagonals $OC$ and $AB$ bisect each other.

1.The midpoint of $OC$ has position vector $\frac12[\mathbf{0}+(\mathbf{a}+\mathbf{b})]=\frac12(\mathbf{a}+\mathbf{b})$ .
2.The midpoint of $AB$ has position vector $\frac12(\mathbf{a}+\mathbf{b})$ as well.
3.Since the two diagonals have the same midpoint, each bisects the other.

Answer: Both diagonals have midpoint position vector $\frac12(\mathbf{a}+\mathbf{b})$ , so they bisect each other.

Common mistakes

Don't assume collinearity from proportional-looking coordinates without finding one common scalar multiplier.
Don't use a diagram alone as proof and never shows that the two candidate midpoints have the same position vector.

Exam tip

In a vector proof, calculate both relevant position vectors and use their equality to justify the geometric conclusion.

Tier 1 · Easy

1.
Two forces acting on a particle are $\mathbf{F}_1=(4,-1)$ N and $\mathbf{F}_2=(-2,5)$ N. Find the resultant force and its magnitude.
(3)
(Total for Question 1 is 3 marks)
2.
Three forces acting on a particle are $(5,-2)$ N, $(-1,6)$ N and $(-4,-4)$ N. Determine whether the particle is in equilibrium, showing the component calculation.
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1.
Two forces acting on a particle are $\mathbf{F}_1=(4,-1)\,\text{N}$ and $\mathbf{F}_2=(-2,5)\,\text{N}$ . Find the third force required for equilibrium and its magnitude.
(3)
(Total for Question 1 is 3 marks)
2.
Forces $(7,-4)$ N and $(-1,10)$ N act on a particle. A third force has the form $\lambda(-1,-1)$ N. Find $\lambda$ so that the particle is in equilibrium, and find the exact magnitude of the third force.
(4)
(Total for Question 2 is 4 marks)
3.
A survey vehicle travels $12$ km east and $5$ km north, followed by $4$ km west and $2$ km south. Find the vector for the direct return journey, taking east and north as the positive component directions. Find the return distance and the bearing of the return journey, giving the distance to $3$ significant figures and the bearing to the nearest degree.
(5)
(Total for Question 3 is 5 marks)

Tier 3 · Hard

1.
Three vectors sum to the zero vector. Two of them are $\mathbf{a}=(3,4,-2)$ and $\mathbf{b}=(-5,1,6)$ . Find the third vector $\mathbf{c}$ , its exact magnitude and a unit vector in its direction.
(5)
(Total for Question 1 is 5 marks)
2.
A particle is held in equilibrium by two tensions and its weight. The tensions are $P(\tfrac35,\tfrac45)$ N and $Q(-\tfrac5{13},\tfrac{12}{13})$ N, where $P>0$ and $Q>0$ , and the weight is $(0,-56)$ N. Find $P$ and $Q$ , showing both component equations and checking the equilibrium.
(6)
(Total for Question 2 is 6 marks)
3.
A boat moves relative to the water at $12$ km h $^{-1}$ on a bearing of $030^\circ$ . A current of $4$ km h $^{-1}$ acts due east. Find the boat's ground-velocity vector in east-north component order. Hence find its ground speed, its bearing to the nearest $0.1^\circ$ , and the time taken to travel $25$ km in the direction of this ground velocity. Give the speed and time to $3$ significant figures, using unrounded values in your working.
(6)
(Total for Question 3 is 6 marks)
4.
Two survey drones move in a horizontal coordinate plane, with distances in kilometres and time $t$ in hours. Drone $A$ starts at $(0,0)$ with velocity $(2,1)$ , while drone $B$ starts at $(10,-2)$ with velocity $(-1,3)$ . Find the displacement of $B$ from $A$ at time $t$ . Hence find the first time at which the drones are $5$ km apart, giving the time to $3$ significant figures.
(6)
(Total for Question 4 is 6 marks)
5.
An aircraft must travel due north through a wind of $60$ km h $^{-1}$ due east. Its speed relative to the air is $250$ km h $^{-1}$ . Let $\mathbf{i}$ point east and $\mathbf{j}$ point north. Find the required air-velocity vector, the bearing on which the aircraft must head to the nearest $0.1^\circ$ , and its ground speed to $3$ significant figures.
(6)
(Total for Question 5 is 6 marks)

Answer key

Answers begin on a new printed page so the question pack can be completed without the solutions alongside it.

10.1 · Use vectors in two dimensions and in three dimensions.

Tier 1 · Easy

Mark scheme for 10.1 Tier 1 · Easy
Question	Scheme	Marks
1	$\overrightarrow{AB}=(6,-4)$	2
(2 marks)		2
Notes
Subtract the coordinates of $A$ from those of $B$ : $\overrightarrow{AB}=(4-(-2),1-5)=(6,-4)$ .
2	$\mathbf{a}=(-4,7,-2)$ and $3\mathbf{a}=(-12,21,-6)$	2
(2 marks)		2
Notes
The coefficients of $\mathbf{i}$ , $\mathbf{j}$ and $\mathbf{k}$ are the three components, so $\mathbf{a}=(-4,7,-2)$ . Multiplying every component by $3$ gives $3\mathbf{a}=(-12,21,-6)$ .

Tier 2 · Standard

Mark scheme for 10.1 Tier 2 · Standard
Question	Scheme	Marks
1	$\lambda=1$ , $\mu=2$	4
(4 marks)		4
Notes
Comparing the first two components gives $\lambda+2\mu=5$ and $2\lambda-\mu=0$ . The second equation gives $\mu=2\lambda$ , so the first gives $5\lambda=5$ and $\lambda=1$ , $\mu=2$ . The third component checks: $-\lambda+3\mu=-1+6=5$ .
2	$B=(-1,4,6)$ and $\overrightarrow{BA}=(3,-5,-2)$ ; their sum is $(0,0,0)$ .	3
(3 marks)		3
Notes
Add the displacement to the coordinates of $A$ : $B=(2,-1,4)+(-3,5,2)=(-1,4,6)$ . Reversing a displacement changes its sign, so $\overrightarrow{BA}=(3,-5,-2)$ . Componentwise addition gives $(-3,5,2)+(3,-5,-2)=(0,0,0)$ .
3	$p=3$ , $q=3$ , $r=5$ and $2\mathbf{a}-\mathbf{b}=(2,13,-11)$	4
(4 marks)		4
Notes
Equality of corresponding components gives $p+4=7$ , $2q-1=5$ and $-3+r=2$ , so $p=3$ , $q=3$ and $r=5$ . Hence $\mathbf{a}=(3,6,-3)$ and $\mathbf{b}=(4,-1,5)$ , giving $2\mathbf{a}-\mathbf{b}=(6,12,-6)-(4,-1,5)=(2,13,-11)$ .

Tier 3 · Hard

Mark scheme for 10.1 Tier 3 · Hard
Question	Scheme	Marks
1	The spanning condition is $u+v-w=0$ . $(7,1,k)$ requires $k=8$ , whereas $(5,k,7)$ requires $k=2$ ; no single value of $k$ satisfies both.	4
(4 marks)		4
Notes
A linear combination has components $(u,v,w)=(\lambda+2\mu,\lambda-\mu,2\lambda+\mu)$ , so $u+v-w=0$ . Conversely, if $u+v-w=0$ , choose $\lambda=(u+2v)/3$ and $\mu=(u-v)/3$ ; these give the stated first two components and $2\lambda+\mu=u+v=w$ . Thus the condition is necessary and sufficient. For $(7,1,k)$ it gives $7+1-k=0$ , so $k=8$ . For $(5,k,7)$ it gives $5+k-7=0$ , so $k=2$ . Since $8\ne2$ , there is no common value.
2	$\lambda=1$ , the final point is $(5,-2,0)$ and the single displacement is $(4,0,-5)$ .	5
(5 marks)		5
Notes
The final $z$ -coordinate is $5-2+\lambda-4=\lambda-1$ . Since it is zero, $\lambda=1$ . The total displacement is $(3,1,-2)+(-1,2,1)+(2,-3,-4)=(4,0,-5)$ . Adding this to $A$ gives the final point $(1,-2,5)+(4,0,-5)=(5,-2,0)$ .
3	$\mathbf{a}=(3,-2,4)$ and $\mathbf{b}=(1,5,-3)$ Check: $2(3,-2,4)+(1,5,-3)=(7,1,5)$ and $(3,-2,4)-2(1,5,-3)=(1,-12,10)$ .	4
(4 marks)		4
Notes
Write $\mathbf{u}=(7,1,5)$ and $\mathbf{v}=(1,-12,10)$ . Doubling $2\mathbf{a}+\mathbf{b}=\mathbf{u}$ and adding $\mathbf{a}-2\mathbf{b}=\mathbf{v}$ gives $5\mathbf{a}=2\mathbf{u}+\mathbf{v}=(15,-10,20)$ , so $\mathbf{a}=(3,-2,4)$ . Then $\mathbf{b}=\mathbf{u}-2\mathbf{a}=(1,5,-3)$ . Checking, $2\mathbf{a}+\mathbf{b}=(7,1,5)$ and $\mathbf{a}-2\mathbf{b}=(1,-12,10)$ .
4	$p=3$ , $q=5$ and $r=4$ .	5
(5 marks)		5
Notes
Comparing components in $p\mathbf{u}+q\mathbf{v}+r\mathbf{w}=(p+r,p+q,q+r)$ gives $p+r=7$ , $p+q=8$ and $q+r=9$ . Adding the first two equations and subtracting the third gives $2p=6$ , so $p=3$ . It follows that $q=5$ and $r=4$ . The check is $3(1,1,0)+5(0,1,1)+4(1,0,1)=(3+4,3+5,5+4)=(7,8,9)$ .
5	$\mathbf{a}=(4,3,6)$ , $\mathbf{b}=(3,6,4)$ and $\mathbf{c}=(6,4,3)$ .	5
(5 marks)		5
Notes
The sum equation gives $p+q+r=13$ . The difference equation gives $p-q=1$ , $q-r=-3$ and $r-p=2$ . Thus $p=q+1$ and $r=q+3$ , so $(q+1)+q+(q+3)=13$ and $q=3$ . Hence $p=4$ and $r=6$ . Therefore $\mathbf{a}=(4,3,6)$ , $\mathbf{b}=(3,6,4)$ and $\mathbf{c}=(6,4,3)$ . Their sum is $(13,13,13)$ and $\mathbf{a}-\mathbf{b}=(1,-3,2)$ , as required.

10.2 · Calculate the magnitude and direction of a vector and convert between component form and magnitude/direction form.

Tier 1 · Easy

Mark scheme for 10.2 Tier 1 · Easy
Question	Scheme	Marks
1	Magnitude $5$ ; direction $53.1^\circ$	3
(3 marks)		3
Notes
The magnitude is $\sqrt{3^2+4^2}=5$ . Both components are positive, so the vector is in the first quadrant and $\theta=\tan^{-1}(4/3)=53.1^\circ$ .
2	$\left(-\dfrac45,\dfrac35\right)$	2
(2 marks)		2
Notes
The magnitude of $(-8,6)$ is $\sqrt{64+36}=10$ . Dividing both components by $10$ gives the unit vector $(-8/10,6/10)=(-4/5,3/5)$ .

Tier 2 · Standard

Mark scheme for 10.2 Tier 2 · Standard
Question	Scheme	Marks
1	Magnitude $10$ Direction $127^\circ$ to $3$ significant figures	3
(3 marks)		3
Notes
The magnitude is $\sqrt{(-6)^2+8^2}=\sqrt{100}=10$ . The vector lies in quadrant II. Its reference angle is $\arctan(8/6)=53.130\ldots^\circ$ , so its direction is $180^\circ-53.130\ldots^\circ=126.870\ldots^\circ$ .
2	$(6,-6\sqrt3)$	3
(3 marks)		3
Notes
For a bearing measured from north, the east component is $12\sin150^\circ=6$ and the north component is $12\cos150^\circ=-6\sqrt3$ . Hence the component form, in east-north order, is $(6,-6\sqrt3)$ .
3	The vector is $(-4\sqrt5,-2\sqrt5)$ and its direction angle is $206.6^\circ$ .	4
(4 marks)		4
Notes
Write the components as $(-2k,-k)$ with $k>0$ . The magnitude condition gives $\sqrt{4k^2+k^2}=k\sqrt5=10$ , so $k=2\sqrt5$ and the vector is $(-4\sqrt5,-2\sqrt5)$ . It lies in the third quadrant and has reference angle $\tan^{-1}(1/2)=26.565\ldots^\circ$ , so its direction is $180^\circ+26.565\ldots^\circ=206.6^\circ$ .

Tier 3 · Hard

Mark scheme for 10.2 Tier 3 · Hard
Question	Scheme	Marks
1	$(5,12)$ or $(5,-12)$	4
(4 marks)		4
Notes
The horizontal component is $13(5/13)=5$ . If the vertical component is $y$ , then $5^2+y^2=13^2$ , so $y^2=144$ and $y=\pm12$ . Cosine fixes the positive horizontal component but does not distinguish an angle above the axis from its reflection below the axis.
2	$k=-2$ and the direction angle is $257.5^\circ$	5
(5 marks)		5
Notes
$k^2+(k-7)^2=85$ gives $2k^2-14k-36=0$ , so $(k-9)(k+2)=0$ and $k=9$ or $k=-2$ . The stated third-quadrant direction selects $k=-2$ , giving $\mathbf{v}=(-2,-9)$ . Its reference angle is $\tan^{-1}(9/2)=77.471\ldots^\circ$ , so the direction is $180^\circ+77.471\ldots^\circ=257.5^\circ$ to $1$ decimal place.
3	$\mathbf{a}+\mathbf{b}=(\sqrt3,7)$ , with magnitude $2\sqrt{13}$ and direction $76.1^\circ$ .	5
(5 marks)		5
Notes
$\mathbf{a}=(8\cos30^\circ,8\sin30^\circ)=(4\sqrt3,4)$ and $\mathbf{b}=(6\cos150^\circ,6\sin150^\circ)=(-3\sqrt3,3)$ . Hence $\mathbf{a}+\mathbf{b}=(\sqrt3,7)$ . Its magnitude is $\sqrt{3+49}=2\sqrt{13}$ . Both components are positive, so its direction is $\tan^{-1}(7/\sqrt3)=76.102\ldots^\circ=76.1^\circ$ to $1$ decimal place.
4	$\mathbf{b}=(3,-4)$ , $\|\mathbf{a}+\mathbf{b}\|=6$ , and the direction angle of $\mathbf{b}$ is $306.9^\circ$ .	5
(5 marks)		5
Notes
For the resultant to be horizontal, the vertical component of $\mathbf{b}$ must be $-4$ . Write $\mathbf{b}=(x,-4)$ . Since $\|\mathbf{b}\|=5$ , $x^2+16=25$ , so $x=\pm3$ . If $x=-3$ , the resultant is the zero vector and does not point along the positive $x$ -axis. Therefore $\mathbf{b}=(3,-4)$ and $\mathbf{a}+\mathbf{b}=(6,0)$ , whose magnitude is $6$ . The direction of $\mathbf{b}$ is $360^\circ-\tan^{-1}(4/3)=306.869\ldots^\circ$ , giving $306.9^\circ$ .
5	The direction angle of $\mathbf{v}$ is $75^\circ$ , and $\mathbf{v}=(2\sqrt6-2\sqrt2,2\sqrt6+2\sqrt2)$ .	5
(5 marks)		5
Notes
The direction angles of $\mathbf{a}$ and $\mathbf{b}$ are $30^\circ$ and $120^\circ$ , respectively. Their smaller angle is $90^\circ$ , so its internal bisector has direction angle $(30^\circ+120^\circ)/2=75^\circ$ . Hence $\mathbf{v}=(8\cos75^\circ,8\sin75^\circ)$ . Using $\cos75^\circ=(\sqrt6-\sqrt2)/4$ and $\sin75^\circ=(\sqrt6+\sqrt2)/4$ gives $\mathbf{v}=(2\sqrt6-2\sqrt2,2\sqrt6+2\sqrt2)$ .

10.3 · Add vectors diagrammatically and perform the algebraic operations of vector addition and multiplication by scalars, and understand their geometrical interpretations.

Tier 1 · Easy

Mark scheme for 10.3 Tier 1 · Easy
Question	Scheme	Marks
1	$(-3,1)$	1
(1 mark)		1
Notes
Add corresponding components: $\mathbf{u}+\mathbf{v}=(2+(-5),-3+4)=(-3,1)$ .
2	$\mathbf{w}=-2\mathbf{u}$ , so it points in the opposite direction and has twice the magnitude of $\mathbf{u}$ .	2
(2 marks)		2
Notes
Multiplying both components of $\mathbf{u}$ by $-2$ gives $(-6,2)=\mathbf{w}$ . A negative scalar reverses direction, and its absolute value $2$ doubles the magnitude.

Tier 2 · Standard

Mark scheme for 10.3 Tier 2 · Standard
Question	Scheme	Marks
1	$\overrightarrow{AM}=\dfrac12(\mathbf{u}+\mathbf{v})$	3
(3 marks)		3
Notes
Take $A$ as the origin. Then the position vectors of $B$ and $C$ are $\mathbf{u}$ and $\mathbf{v}$ . The midpoint has position vector equal to their average, so $\overrightarrow{AM}=\tfrac12(\mathbf{u}+\mathbf{v})$ .
2	The resultant is $(-1,2)$ and the return displacement is $(1,-2)$ ; placed head-to-tail, the return vector closes the path.	3
(3 marks)		3
Notes
Add corresponding components: $(4,-1)+(-2,5)+(-3,-2)=(-1,2)$ . The displacement back to the start is the negative of this resultant, $(1,-2)$ . Drawing this return vector from the final head to the initial tail makes the four-vector path closed.
3	$\mathbf{a}-\mathbf{b}=(4,5)=\overrightarrow{BA}$	3
(3 marks)		3
Notes
$\mathbf{a}-\mathbf{b}=(5,2)-(1,-3)=(4,5)$ . Since $\overrightarrow{BA}=\overrightarrow{OA}-\overrightarrow{OB}$ , the difference is the directed segment from $B$ to $A$ .

Tier 3 · Hard

Mark scheme for 10.3 Tier 3 · Hard
Question	Scheme	Marks
1	$\alpha=2$ and $\beta=3$ , so $\mathbf{w}=2\mathbf{a}+3\mathbf{b}$ .	4
(4 marks)		4
Notes
Equating components gives $\alpha+3\beta=11$ and $2\alpha-\beta=1$ . From the second, $\beta=2\alpha-1$ ; substitution gives $7\alpha=14$ , hence $\alpha=2$ and $\beta=3$ . Geometrically, two copies of $\mathbf{a}$ and three copies of $\mathbf{b}$ placed head-to-tail have resultant $\mathbf{w}$ .
2	$\lambda=-2$ and $\mathbf{r}=2(1,3)=(2,6)$ ; the resultant is parallel to $(1,3)$ , in the same direction and twice its magnitude.	4
(4 marks)		4
Notes
$\mathbf{r}=(4+\lambda,2-2\lambda)$ . If $\mathbf{r}=k(1,3)$ , then $4+\lambda=k$ and $2-2\lambda=3k$ . Eliminating $k$ gives $2-2\lambda=12+3\lambda$ , so $\lambda=-2$ and $k=2$ . Thus $\mathbf{r}=(2,6)=2(1,3)$ , which proves the stated parallel, same-direction relationship and the factor-two scaling.
3	$\overrightarrow{DE}=\dfrac23(\mathbf{b}-\mathbf{a})=\dfrac23\overrightarrow{BC}$ , so $DE$ is parallel to $BC$ and $DE:BC=2:3$ .	5
(5 marks)		5
Notes
$\overrightarrow{AD}=\frac23\mathbf{a}$ and $\overrightarrow{AE}=\frac23\mathbf{b}$ . Therefore $\overrightarrow{DE}=\overrightarrow{AE}-\overrightarrow{AD}=\frac23(\mathbf{b}-\mathbf{a})$ . Also $\overrightarrow{BC}=\overrightarrow{AC}-\overrightarrow{AB}=\mathbf{b}-\mathbf{a}$ , so $\overrightarrow{DE}=\frac23\overrightarrow{BC}$ . The positive scalar multiple proves that the segments are parallel in the same direction, and their lengths are in the ratio $2:3$ .
4	$\overrightarrow{ON}=\dfrac13(\mathbf{a}+\mathbf{b})$ ; $A,N,P$ are collinear and $AN:NP=2:1$ .	6
(6 marks)		6
Notes
Since $M$ is the midpoint of $AB$ , $\overrightarrow{OM}=(\mathbf{a}+\mathbf{b})/2$ . The ratio $ON:NM=2:1$ gives $\overrightarrow{ON}=(2/3)\overrightarrow{OM}=(\mathbf{a}+\mathbf{b})/3$ . Also $\overrightarrow{OP}=\mathbf{b}/2$ . Therefore $\overrightarrow{AN}=\overrightarrow{ON}-\overrightarrow{OA}=(\mathbf{b}-2\mathbf{a})/3$ , while $\overrightarrow{NP}=\overrightarrow{OP}-\overrightarrow{ON}=(\mathbf{b}-2\mathbf{a})/6$ . Thus $\overrightarrow{AN}=2\overrightarrow{NP}$ , proving that $A,N,P$ are collinear in that order and $AN:NP=2:1$ .
5	$\overrightarrow{PQ}=\overrightarrow{SR}=\dfrac12(\mathbf{c}-\mathbf{a})$ and $\overrightarrow{QR}=\overrightarrow{PS}=\dfrac12(\mathbf{d}-\mathbf{b})$ , so $PQRS$ is a parallelogram.	5
(5 marks)		5
Notes
The midpoint position vectors are $(\mathbf{a}+\mathbf{b})/2$ , $(\mathbf{b}+\mathbf{c})/2$ , $(\mathbf{c}+\mathbf{d})/2$ and $(\mathbf{d}+\mathbf{a})/2$ . Hence $\overrightarrow{PQ}=(\mathbf{c}-\mathbf{a})/2$ and $\overrightarrow{SR}=(\mathbf{c}-\mathbf{a})/2$ . Similarly, $\overrightarrow{QR}=(\mathbf{d}-\mathbf{b})/2$ and $\overrightarrow{PS}=(\mathbf{d}-\mathbf{b})/2$ . Both pairs of opposite directed sides are equal, so $PQRS$ is a parallelogram.

10.4 · Understand and use position vectors; calculate the distance between two points represented by position vectors.

Tier 1 · Easy

Mark scheme for 10.4 Tier 1 · Easy
Question	Scheme	Marks
1	$\overrightarrow{AB}=(5,4)$ and $AB=\sqrt{41}$	3
(3 marks)		3
Notes
Subtract the position vector of $A$ from that of $B$ : $\overrightarrow{AB}=(7-2,3-(-1))=(5,4)$ . Therefore $AB=\sqrt{5^2+4^2}=\sqrt{41}$ .
2	$\overrightarrow{OB}=(-2,2,3)$	2
(2 marks)		2
Notes
$\overrightarrow{OB}=\overrightarrow{OA}+\overrightarrow{AB}=(3,-2,1)+(-5,4,2)=(-2,2,3)$ .

Tier 2 · Standard

Mark scheme for 10.4 Tier 2 · Standard
Question	Scheme	Marks
1	Midpoint $(2,3,1)$ $P=(4,4,-1)$	4
(4 marks)		4
Notes
The midpoint is the componentwise average: $((-2+6)/2,(1+5)/2,(5-3)/2)=(2,3,1)$ . Also $\overrightarrow{AB}=(8,4,-8)$ . Since $AP$ is three quarters of $AB$ , $\overrightarrow{OP}=(-2,1,5)+\tfrac34(8,4,-8)=(4,4,-1)$ .
2	$t=3$ and $AP=2\sqrt3$	4
(4 marks)		4
Notes
$AP^2=(t-1)^2+(0-2)^2+(1-(-1))^2=(t-1)^2+8$ , while $BP^2=(t-5)^2+(0+2)^2+(1-3)^2=(t-5)^2+8$ . Equating these gives $(t-1)^2=(t-5)^2$ , hence $t=3$ . Then $AP=\sqrt{(3-1)^2+8}=\sqrt{12}=2\sqrt3$ .
3	$t=-1$ or $t=7$	4
(4 marks)		4
Notes
$\overrightarrow{AB}=(3-t,-4,6)$ , so $AB^2=(3-t)^2+16+36$ . Since $AB=2\sqrt{17}$ , $(3-t)^2+52=68$ , giving $(3-t)^2=16$ . Hence $3-t=\pm4$ , so $t=-1$ or $t=7$ .

Tier 3 · Hard

Mark scheme for 10.4 Tier 3 · Hard
Question	Scheme	Marks
1	$\overrightarrow{OM}=(7,4,0)$ and $OM=\sqrt{65}$	5
(5 marks)		5
Notes
Since $AM$ is three of the four equal ratio parts, $\overrightarrow{OM}=\overrightarrow{OA}+\frac34(\overrightarrow{OB}-\overrightarrow{OA})$ . Now $\overrightarrow{OB}-\overrightarrow{OA}=(8,8,-4)$ , so $\overrightarrow{OM}=(1,-2,3)+(6,6,-3)=(7,4,0)$ . Thus $OM=\sqrt{7^2+4^2}=\sqrt{65}$ .
2	$AB=AC=3$ and $BC=3\sqrt2$ , so the triangle is right-angled at $A$ . $\overrightarrow{OM}=(1,\tfrac32,\tfrac72)$ and $MA=MB=MC=\dfrac{3\sqrt2}{2}$ .	6
(6 marks)		6
Notes
$\overrightarrow{AB}=(2,1,2)$ and $\overrightarrow{AC}=(-2,2,1)$ , so $AB=AC=\sqrt9=3$ . Also $\overrightarrow{BC}=(-4,1,-1)$ , so $BC=\sqrt{18}=3\sqrt2$ . Since $AB^2+AC^2=9+9=18=BC^2$ , the right angle is at $A$ and $BC$ is the hypotenuse. Its midpoint is $M=((3-1)/2,(1+2)/2,(4+3)/2)=(1,3/2,7/2)$ . Now $\overrightarrow{AM}=(0,3/2,3/2)$ has magnitude $3\sqrt2/2$ , while $MB=MC=BC/2=3\sqrt2/2$ .
3	$\overrightarrow{OP}=(7,10,-9)$ , with $AP=18$ and $PB=9$ , so $AP:PB=2:1$ .	5
(5 marks)		5
Notes
$\overrightarrow{AB}=(3,6,-6)$ . Since $P$ lies beyond $B$ and $AP=2PB$ , $\overrightarrow{AP}=2\overrightarrow{AB}=(6,12,-12)$ . Hence $\overrightarrow{OP}=\overrightarrow{OA}+\overrightarrow{AP}=(7,10,-9)$ . Now $AP=\sqrt{6^2+12^2+(-12)^2}=18$ , while $\overrightarrow{BP}=(3,6,-6)$ gives $PB=9$ . Thus the required ratio is verified.
4	$\overrightarrow{AB}=(3,4,2)$ and $\overrightarrow{BC}=2\overrightarrow{AB}$ , so the points are collinear with $B$ between $A$ and $C$ . $AB=\sqrt{29}$ , $BC=2\sqrt{29}$ , $AC=3\sqrt{29}$ , so $AB:BC=1:2$ .	5
(5 marks)		5
Notes
$\overrightarrow{AB}=(4,6,1)-(1,2,-1)=(3,4,2)$ and $\overrightarrow{BC}=(10,14,5)-(4,6,1)=(6,8,4)=2\overrightarrow{AB}$ . The positive scalar multiple shows that the directions agree, so $A,B,C$ are collinear with $B$ between the other two points. The distances are $AB=\sqrt{3^2+4^2+2^2}=\sqrt{29}$ , $BC=\sqrt{6^2+8^2+4^2}=2\sqrt{29}$ and $AC=\sqrt{9^2+12^2+6^2}=3\sqrt{29}$ . Therefore $AB:BC=1:2$ .
5	$k=\pm\sqrt{11}$ gives $AB=AC$ , and $k=\pm\sqrt3$ gives $AB=BC$ .	6
(6 marks)		6
Notes
The squared side lengths are $AB^2=16$ , $AC^2=1+k^2+4=k^2+5$ and $BC^2=(-3)^2+k^2+2^2=k^2+13$ . The condition $AB=AC$ gives $16=k^2+5$ , so $k=\pm\sqrt{11}$ . The condition $AB=BC$ gives $16=k^2+13$ , so $k=\pm\sqrt3$ . The equation $AC=BC$ would require $k^2+5=k^2+13$ , which is impossible. These four values are therefore all the possibilities.

10.5 · Use vectors to solve problems in pure mathematics and in context (including forces).

Tier 1 · Easy

Mark scheme for 10.5 Tier 1 · Easy
Question	Scheme	Marks
1	Resultant $(2,4)$ N; magnitude $2\sqrt5$ N	3
(3 marks)		3
Notes
Add the force components: $\mathbf{R}=\mathbf{F}_1+\mathbf{F}_2=(4-2,-1+5)=(2,4)$ N. Its magnitude is $\|\mathbf{R}\|=\sqrt{2^2+4^2}=\sqrt{20}=2\sqrt5$ N.
2	The particle is in equilibrium because the resultant force is $(0,0)$ N.	2
(2 marks)		2
Notes
The resultant is $(5,-2)+(-1,6)+(-4,-4)=(5-1-4,-2+6-4)=(0,0)$ N. A zero resultant is the condition for equilibrium.

Tier 2 · Standard

Mark scheme for 10.5 Tier 2 · Standard
Question	Scheme	Marks
1	$\mathbf{F}_3=(-2,-4)\,\text{N}$ $\|\mathbf{F}_3\|=2\sqrt5\,\text{N}$	3
(3 marks)		3
Notes
For equilibrium the vector sum is zero. The first two forces sum to $(4,-1)+(-2,5)=(2,4)$ , so the third force must be its negative, $(-2,-4)\,\text{N}$ . Its magnitude is $\sqrt{(-2)^2+(-4)^2}=\sqrt{20}=2\sqrt5\,\text{N}$ .
2	$\lambda=6$ and the third force has magnitude $6\sqrt2$ N	4
(4 marks)		4
Notes
The first two forces have resultant $(6,6)$ N. Equilibrium requires $(6,6)+\lambda(-1,-1)=(0,0)$ , so $\lambda=6$ and the third force is $(-6,-6)$ N. Its magnitude is $\sqrt{(-6)^2+(-6)^2}=6\sqrt2$ N.
3	The direct return vector is $(-8,-3)$ km, the distance is $8.54$ km and the bearing is $249^\circ$ .	5
(5 marks)		5
Notes
The outward displacement is $(12,5)+(-4,-2)=(8,3)$ km, so the direct return vector is $(-8,-3)$ km. Its magnitude is $\sqrt{8^2+3^2}=\sqrt{73}=8.544\ldots$ km, giving $8.54$ km to $3$ significant figures. Measured clockwise from north, the return bearing is $180^\circ+\tan^{-1}(8/3)=249.443\ldots^\circ$ , which is $249^\circ$ to the nearest degree.

Tier 3 · Hard

Mark scheme for 10.5 Tier 3 · Hard
Question	Scheme	Marks
1	$\mathbf{c}=(2,-5,-4)$ $\|\mathbf{c}\|=3\sqrt5$ and a unit direction vector is $\frac{1}{3\sqrt5}(2,-5,-4)$ .	5
(5 marks)		5
Notes
The condition $\mathbf{a}+\mathbf{b}+\mathbf{c}=\mathbf{0}$ gives $\mathbf{c}=-(\mathbf{a}+\mathbf{b})$ . The first two vectors sum to $(-2,5,4)$ , so $\mathbf{c}=(2,-5,-4)$ . Its magnitude is $\sqrt{2^2+(-5)^2+(-4)^2}=\sqrt{45}=3\sqrt5$ . Dividing the vector by this magnitude gives the unit vector $\frac{1}{3\sqrt5}(2,-5,-4)$ .
2	$P=25$ N and $Q=39$ N	6
(6 marks)		6
Notes
Horizontal equilibrium gives $3P/5-5Q/13=0$ , so $39P=25Q$ . Vertical equilibrium gives $4P/5+12Q/13-56=0$ . From the first equation $Q=39P/25$ ; substitution into the second gives $4P/5+36P/25=56$ , so $56P/25=56$ and $P=25$ . Hence $Q=39$ . Checking, the two tensions are $(15,20)$ N and $(-15,36)$ N; adding the weight $(0,-56)$ N gives $(0,0)$ N.
3	The ground-velocity vector is $(10,6\sqrt3)$ km h $^{-1}$ . The ground speed is $14.4$ km h $^{-1}$ , the bearing is $043.9^\circ$ , and the time is $1.73$ h.	6
(6 marks)		6
Notes
The boat's velocity relative to the water has east-north components $(12\sin30^\circ,12\cos30^\circ)=(6,6\sqrt3)$ . Adding the current $(4,0)$ gives ground velocity $(10,6\sqrt3)$ km h $^{-1}$ . Its magnitude is $\sqrt{100+108}=4\sqrt{13}=14.422\ldots$ km h $^{-1}$ . The bearing is $\tan^{-1}[10/(6\sqrt3)]=43.897\ldots^\circ$ , written $043.9^\circ$ . Using the exact speed, the travel time is $25/(4\sqrt{13})=1.7334\ldots$ h, so the required values are $14.4$ km h $^{-1}$ and $1.73$ h to $3$ significant figures.
4	The displacement of $B$ from $A$ is $(10-3t,-2+2t)$ . The two possible times are $t=\dfrac{34\pm\sqrt{129}}{13}$ , so the first time is $1.74$ hours to $3$ significant figures.	6
(6 marks)		6
Notes
At time $t$ , the position vectors are $(2t,t)$ for $A$ and $(10-t,-2+3t)$ for $B$ . Subtracting gives the displacement from $A$ to $B$ as $(10-3t,-2+2t)$ . A separation of $5$ km requires $(10-3t)^2+(-2+2t)^2=25$ , which simplifies to $13t^2-68t+79=0$ . Thus $t=[68\pm\sqrt{516}]/26=(34\pm\sqrt{129})/13$ . The first value is $1.741706\ldots$ , giving $1.74$ hours to $3$ significant figures.
5	The required air-velocity vector is $-60\mathbf{i}+10\sqrt{589}\mathbf{j}$ km h $^{-1}$ . The required bearing is $346.1^\circ$ , and the ground speed is $243$ km h $^{-1}$ to $3$ significant figures.	6
(6 marks)		6
Notes
To cancel the wind's eastward component, the aircraft's air velocity must have east component $-60$ . Write it as $-60\mathbf{i}+v\mathbf{j}$ with $v>0$ . Its magnitude is $250$ , so $60^2+v^2=250^2$ and $v=\sqrt{58900}=10\sqrt{589}$ . Adding the wind vector $60\mathbf{i}$ leaves the ground velocity $10\sqrt{589}\mathbf{j}$ , which is due north and has magnitude $242.693\ldots$ km h $^{-1}$ . The heading is $\tan^{-1}(60/(10\sqrt{589}))=13.886\ldots^\circ$ west of north, so the bearing is $360^\circ-13.886\ldots^\circ=346.1^\circ$ .