Pearson Edexcel IAL Mathematics A2 P4.7 Vectors

[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 ]

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

$81\lambda^2+52\lambda-220=0 \Rightarrow(81\lambda-110)(\lambda+2)=0$ $\lambda=-2,\ \frac{110}{81}$
The other possible position vector is
$\overrightarrow{OA} =\begin{pmatrix} 1+8\cdot\frac{110}{81}\\ 2-\frac{110}{81}\\ 5+4\cdot\frac{110}{81} \end{pmatrix} =\begin{pmatrix} \frac{961}{81}\\ \frac{52}{81}\\ \frac{845}{81} \end{pmatrix}$

(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 ]

$\cos\theta=\pm \frac{ \begin{pmatrix}-15\\4\\-3\end{pmatrix}\cdot \begin{pmatrix}8\\-1\\4\end{pmatrix}} {\sqrt{15^2+4^2+3^2}\sqrt{8^2+1^2+4^2}}$ $=\pm\frac{-136}{45\sqrt{10}}$ $\text{Area }OAB=\frac12(5\sqrt{10})(4\sqrt{10})\sin\theta$ $=29.4\text{ (awrt)}$

[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 ]

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

(b)

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

[ 3 ]

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

[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.

[Maximum number: 3]

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 $|\overrightarrow{O A}|=5 \sqrt{10}$

show that at A the parameter $\lambda$ satisfies

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

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

[Maximum number: 2]

Figure 1

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

Find $\overrightarrow{R Q}$

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

[Maximum number: 3]

The line $l_1$ has equation
$\mathbf r=\begin{pmatrix}2\\3\\-7\end{pmatrix}+\lambda\begin{pmatrix}1\\2\\2\end{pmatrix}$
where $\lambda$ is a scalar parameter.

The line $l_2$ has equation
$\mathbf r=\begin{pmatrix}2\\3\\-7\end{pmatrix}+\mu\begin{pmatrix}4\\-1\\8\end{pmatrix}$
where $\mu$ is a scalar parameter.

Given that $l_1$ and $l_2$ meet at the point P.

The point Q lies on $l_1$ where $\lambda=6$ .
Given that point R lies on $l_2$ such that triangle QPR is an isosceles triangle with PQ=PR.

find the coordinates of the possible positions of point R

Attempts a correct method of finding at least one value for $\mu$ , for example
$(4\mu)^2+\mu^2+(8\mu)^2=6^2+12^2+12^2, \quad \mu=\pm2$
Attempts one correct position:
$\mathbf r=\begin{pmatrix}2\\3\\-7\end{pmatrix}\pm2\begin{pmatrix}4\\-1\\8\end{pmatrix}$
Possible coordinates are (10,1,9) and (-6,5,-23).

[Maximum number: 1]

The line $l_1$ has equation
$\mathbf r=\begin{pmatrix}2\\3\\-7\end{pmatrix}+\lambda\begin{pmatrix}1\\2\\2\end{pmatrix}$
where $\lambda$ is a scalar parameter.

The line $l_2$ has equation
$\mathbf r=\begin{pmatrix}2\\3\\-7\end{pmatrix}+\mu\begin{pmatrix}4\\-1\\8\end{pmatrix}$
where $\mu$ is a scalar parameter.

Given that $l_1$ and $l_2$ meet at the point P.

The point Q lies on $l_1$ where $\lambda=6$ .
Given that point R lies on $l_2$ such that triangle QPR is an isosceles triangle with PQ=PR.

state the coordinates of P

[Maximum number: 3]

The line $l_1$ has equation
$\mathbf r=\begin{pmatrix}2\\3\\-7\end{pmatrix}+\lambda\begin{pmatrix}1\\2\\2\end{pmatrix}$
where $\lambda$ is a scalar parameter.

The line $l_2$ has equation
$\mathbf r=\begin{pmatrix}2\\3\\-7\end{pmatrix}+\mu\begin{pmatrix}4\\-1\\8\end{pmatrix}$
where $\mu$ is a scalar parameter.

Given that $l_1$ and $l_2$ meet at the point P.

The point Q lies on $l_1$ where $\lambda=6$ .
Given that point R lies on $l_2$ such that triangle QPR is an isosceles triangle with PQ=PR.

Given that the angle between lines $l_1$ and $l_2$ is $\theta$ ,
find the value of $\cos\theta$ , giving the answer as a fully simplified fraction.

$\begin{pmatrix}1\\2\\2\end{pmatrix}\cdot\begin{pmatrix}4\\-1\\8\end{pmatrix}=1\cdot4+2\cdot(-1)+2\cdot8=18$
Attempts $\mathbf a\cdot\mathbf b=|\mathbf a||\mathbf b|\cos\theta$ :
$18=\sqrt{1^2+2^2+2^2}\sqrt{4^2+(-1)^2+8^2}\cos\theta$ $\cos\theta=\frac23$

Edexcel A-Level Mathematics A2 P4.7 Vectors Questions

Exam points

Covered syllabus points

Covered syllabus points

Question 1

Question (a)

Question (b)

Question (i)

Question (ii)

Question (c)

Question 2

Question (a)

Question (b)

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8