Pearson Edexcel IAL Mathematics P4.7 Vectors Question Bank

Practise vector geometry in two and three dimensions, including lines, distances, angles, areas and perpendicularity tests.

Exam points

calculate vectors, distances and triangle areas from position vectors or coordinates
write vector equations of lines using a point and direction vector
use scalar products to test perpendicularity or find angles in 3D geometry

Covered syllabus points

P4.7.1 - Vectors in two and three dimensions P4.7.2 - Magnitude of a vector P4.7.3 - Vector addition and scalar multiplication P4.7.4 - Position vectors P4.7.5 - distance between two points P4.7.6 - Vector equations of lines P4.7.7 - Scalar product

[Maximum number: 12]

With respect to a fixed origin O, the line $l_{1}$ is given by the equation

\mathbf{r}=\mathbf{i}+2 \mathbf{j}+5 \mathbf{k}+\lambda(8 \mathbf{i}-\mathbf{j}+4 \mathbf{k})

where $\lambda$ is a scalar parameter.
The point A lies on $l_{1}$
Given that $|\overrightarrow{O A}|=5 \sqrt{10}$

(a)

show that at A the parameter $\lambda$ satisfies

81 \lambda^{2}+52 \lambda-220=0

[ 3 ]

$|\overrightarrow{OA}|^2=(1+8\lambda)^2+(2-\lambda)^2+(5+4\lambda)^2$
M1
$|\overrightarrow{OA}|=5\sqrt{10} \Rightarrow (1+8\lambda)^2+(2-\lambda)^2+(5+4\lambda)^2=250$
M1
$\begin{aligned} 64\lambda^2+16\lambda+1+\lambda^2-4\lambda+4+16\lambda^2+40\lambda+25&=250\\ 81\lambda^2+52\lambda-220&=0 \end{aligned}$
A1*

(b)

Hence

[ 5 ]

(i)

show that one possible position vector for A is -15 i+4 j-3 k

[ 2 ]

$81\lambda^2+52\lambda-220=0 \Rightarrow(81\lambda-110)(\lambda+2)=0$ $\lambda=-2,\ \frac{110}{81}$
M1
$\overrightarrow{OA} =\begin{pmatrix}1\\2\\5\end{pmatrix} -2\begin{pmatrix}8\\-1\\4\end{pmatrix} =\begin{pmatrix}-15\\4\\-3\end{pmatrix}$
A1*

(ii)

find the other possible position vector for A.

[ 3 ]

$81\lambda^2+52\lambda-220=0 \Rightarrow(81\lambda-110)(\lambda+2)=0$ $\lambda=-2,\ \frac{110}{81}$
M1

The other possible position vector is
$\overrightarrow{OA} =\begin{pmatrix} 1+8\cdot\frac{110}{81}\\ 2-\frac{110}{81}\\ 5+4\cdot\frac{110}{81} \end{pmatrix} =\begin{pmatrix} \frac{961}{81}\\ \frac{52}{81}\\ \frac{845}{81} \end{pmatrix}$
A1

(c)

The line $l_{2}$ is parallel to $l_{1}$ and passes through O.
Given that
- $\overrightarrow{OA}=-15\mathbf{i}+4\mathbf{j}-3\mathbf{k}$
- point B lies on $l_{2}$ where $|\overrightarrow{OB}|=4\sqrt{10}$

find the area of triangle OAB, giving your answer to one decimal place.

[ 4 ]

$\cos\theta=\pm \frac{ \begin{pmatrix}-15\\4\\-3\end{pmatrix}\cdot \begin{pmatrix}8\\-1\\4\end{pmatrix}} {\sqrt{15^2+4^2+3^2}\sqrt{8^2+1^2+4^2}}$ $=\pm\frac{-136}{45\sqrt{10}}$
M1 A1

$\text{Area }OAB=\frac12(5\sqrt{10})(4\sqrt{10})\sin\theta$ $=29.4\text{ (awrt)}$
M1 A1

[Maximum number: 5]

Figure 1

Figure 1 shows a sketch of triangle PQR.
Given that
- $\overrightarrow{P Q}=2 \mathbf{i}-3 \mathbf{j}+4 \mathbf{k}$
- $\overrightarrow{P R}=8 \mathbf{i}-5 \mathbf{j}+3 \mathbf{k}$

(a)

Find $\overrightarrow{R Q}$

[ 2 ]

$\overrightarrow{RQ}=(2\mathbf{i}-3\mathbf{j}+4\mathbf{k})-(8\mathbf{i}-5\mathbf{j}+3\mathbf{k})$ $\overrightarrow{RQ}=-6\mathbf{i}+2\mathbf{j}+\mathbf{k}$
M1
A1

(b)

Find the size of angle PQR, in degrees, to three significant figures.

[ 3 ]

$\overrightarrow{PQ}\cdot\overrightarrow{RQ}=2(-6)+(-3)(2)+4(1)=-14$ $-14=\sqrt{29}\sqrt{41}\cos\angle PQR$ $\angle PQR=114^\circ$
M1
dM1
A1

[Maximum number: 3]

Relative to a fixed origin O, the points A, B and C have position vector a, b and c respectively.

Points A, B and C lie in a straight line, with B lying between A and C.
Given A B: A C=1: 3 show that

\mathbf{c}=3 \mathbf{b}-2 \mathbf{a}

M1: Attempts any two of $\overrightarrow{A B}, \overrightarrow{A C}$ and $\overrightarrow{B C}$ .
Condone the wrong way around but it must be subtraction.
Allow marked in the correct place on a diagram
dM1: Uses the given information.
Accept $\overrightarrow{A B}=\frac{1}{3} \overrightarrow{A C}, \quad \overrightarrow{B C}=2 \times \overrightarrow{A B}, \quad \overrightarrow{B C}=\frac{2}{3} \times \overrightarrow{A C} \quad$ etc condoning slips as in previous M1.
A1*: Fully correct work inc bracketing leading to the given answer c=3 b-2 a
Expect to see the brackets multiplied out. So $\mathbf{c}-\mathbf{b}=2 \times(\mathbf{b}-\mathbf{a}) \Rightarrow \mathbf{c}-\mathbf{b}=2 \mathbf{b}-2 \mathbf{a} \Rightarrow \mathbf{c}=3 \mathbf{b}-2 \mathbf{a}$ is fine.

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Covered syllabus points

P4.7 - Vectors question 1

Question (a)

Question (b)

Question (i)

Question (ii)

Question (c)

P4.7 - Vectors question 2

Question (a)

Question (b)

P4.7 - Vectors question 3