Edexcel A-Level Mathematics A2 Fp3 5 Vectors Questions

Practise vector methods for lines, planes, areas, volumes, intersections, angles and distances in three-dimensional coordinate geometry.

Exam points

use vector products and scalar triple products for perpendicular vectors, areas and volumes
convert between line and plane forms to find intersections or shortest distances
calculate angles between lines or planes using direction vectors and normals

Covered syllabus points

FP3.5.1 - Vector product and scalar triple product FP3.5.2 - Use of vectors in problems FP3.5.3 - Equation of a plane

[Maximum number: 10]

The skew lines $l_{1}$ and $l_{2}$ have equations

l_{1}: \mathbf{r}=(\mathbf{i}+2 \mathbf{j}-5 \mathbf{k})+\lambda(5 \mathbf{i}+\mathbf{j})

and

l_{2}: \mathbf{r}=(2 \mathbf{i}-4 \mathbf{j}+4 \mathbf{k})+\mu(8 \mathbf{i}-2 \mathbf{j}+3 \mathbf{k})

where $\lambda$ and $\mu$ are scalar parameters.

(a)

Determine a vector that is perpendicular to both $l_{1}$ and $l_{2}$

[ 2 ]

$(5\mathbf i+\mathbf j)\times(8\mathbf i-2\mathbf j+3\mathbf k)$
or solve
$(u\mathbf i+v\mathbf j+w\mathbf k)\cdot(5\mathbf i+\mathbf j)=0, \quad (u\mathbf i+v\mathbf j+w\mathbf k)\cdot(8\mathbf i-2\mathbf j+3\mathbf k)=0$ $5u+v=0,\quad 8u-2v+3w=0$ $\mathbf n=3\mathbf i-15\mathbf j-18\mathbf k$
or any non-zero multiple of $\mathbf i-5\mathbf j-6\mathbf k$ .

(b)

Determine an equation of the plane parallel to $l_{1}$ that contains $l_{2}$

[ 3 ]

(i)

in the form r=a+s b+t c

[ 1 ]

$\mathbf r=(2\mathbf i-4\mathbf j+4\mathbf k) +s(8\mathbf i-2\mathbf j+3\mathbf k) +t(5\mathbf i+\mathbf j)$

(ii)

in the form r.n =p

[ 2 ]

$(2\mathbf i-4\mathbf j+4\mathbf k)\cdot(3\mathbf i-15\mathbf j-18\mathbf k)=-6$ $\mathbf r\cdot(3\mathbf i-15\mathbf j-18\mathbf k)=-6$
or
$\mathbf r\cdot(-\mathbf i+5\mathbf j+6\mathbf k)=2$

(c)

Determine the shortest distance between $l_{1}$ and $l_{2}$

Give your answer in simplest form.

[ 5 ]

Way 1:
$\frac{6}{\sqrt{3^2+15^2+18^2}}=\frac{2}{\sqrt{62}}, \quad \frac{21}{\sqrt{62}}$ $d=\frac{2}{\sqrt{62}}+\frac{21}{\sqrt{62}} =\frac{23}{\sqrt{62}} =\frac{23\sqrt{62}}{62}$

OR

Way 2:
$\overrightarrow{AB}=\pm(-\mathbf i+6\mathbf j-9\mathbf k)$ $d=\frac{|\overrightarrow{AB}\cdot\mathbf n|}{|\mathbf n|} =\frac{69}{\sqrt{558}} =\frac{23}{\sqrt{62}}$

[Maximum number: 10]

The plane $\Pi_1$ has vector equation

\mathbf r=\begin{smallmatrix}5\\3\\0\end{smallmatrix}+s\begin{smallmatrix}3\\0\\1\end{smallmatrix}+t\begin{smallmatrix}1\\-2\\2\end{smallmatrix}

where s and t are scalar parameters.

(a)

Determine a Cartesian equation for $\Pi_1$ .

[ 3 ]

\mathbf n=\begin{smallmatrix}3\\0\\1\end{smallmatrix}\times\begin{smallmatrix}1\\-2\\2\end{smallmatrix}=\begin{smallmatrix}2\\-5\\-6\end{smallmatrix}

\begin{smallmatrix}5\\3\\0\end{smallmatrix}\cdot\begin{smallmatrix}2\\-5\\-6\end{smallmatrix}=-5

\mathbf r\cdot\begin{smallmatrix}2\\-5\\-6\end{smallmatrix}=-5

2x-5y-6z+5=0

M1: calculates a normal vector from the two direction vectors.
M1: calculates the scalar product using a point in the plane.
A1: any correct Cartesian equation.

(b)

The plane $\Pi_2$ has vector equation $\mathbf r\cdot\begin{smallmatrix}5\\-2\\3\end{smallmatrix}=1$ .

Determine a vector equation for the line of intersection of $\Pi_1$ and $\Pi_2$ .

Give your answer in the form $\mathbf r=\mathbf a+\lambda\mathbf b$ , where $\mathbf a$ and $\mathbf b$ are constant vectors and $\lambda$ is a scalar parameter.

[ 4 ]

From part (a), $\Pi_1:2x-5y-6z=-5$ , and $\Pi_2:5x-2y+3z=1$ .

Taking y=0,

2x-6z=-5,\quad 5x+3z=1

x=-\frac14,\quad z=\frac34

A point on the line is $(-\frac14,0,\frac34)$ . A direction vector is

\begin{smallmatrix}2\\-5\\-6\end{smallmatrix}\times\begin{smallmatrix}5\\-2\\3\end{smallmatrix}=\begin{smallmatrix}-27\\-36\\21\end{smallmatrix}

so an equivalent direction vector is $\begin{smallmatrix}9\\12\\-7\end{smallmatrix}$ . Therefore

\mathbf r=\begin{smallmatrix}-\frac14\\0\\\frac34\end{smallmatrix}+\lambda\begin{smallmatrix}9\\12\\-7\end{smallmatrix}

M1/dM1: finds a point on the line of intersection.
ddM1: forms a direction vector from the plane normals.
A1: any correct vector equation.

(c)

The plane $\Pi_3$ has Cartesian equation 4x-3y-z=0.

Use the answer to part (b) to determine the coordinates of the point of intersection of $\Pi_1$ , $\Pi_2$ and $\Pi_3$ .

[ 3 ]

Using the line from part (b),

\mathbf r=\begin{smallmatrix}-\frac14\\0\\\frac34\end{smallmatrix}+\lambda\begin{smallmatrix}9\\12\\-7\end{smallmatrix}

x=-\frac14+9\lambda,\quad y=12\lambda,\quad z=\frac34-7\lambda

Substitute into 4x-3y-z=0:

4\left(-\frac14+9\lambda\right)-3(12\lambda)-\left(\frac34-7\lambda\right)=0

7\lambda-\frac74=0\Rightarrow \lambda=\frac14

(x,y,z)=(2,3,-1)

M1: substitutes the parametric line from part (b) into Pi_3 and solves for the parameter.
dM1: substitutes the parameter into the line.
A1: correct point only.

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Covered syllabus points

Question 1

Question (a)

Question (b)

Question (i)

Question (ii)

Question (c)

Question 2

Question (a)

Question (b)

Question (c)