2021 H2 Maths Paper 1 Q8 — Worked Solution

Part (a)

4 marks

Approach

a plane containing $l_{1}$ and the point $P$ is fixed once we have two directions lying in it — the line’s direction $\mathbf{d}_{1}$ and $\overrightarrow{AP}$ , where $A$ is a point on $l_{1}$ — so we find the normal as their cross product.

$l_{1}$ has direction $\mathbf{d}_{1} = \begin{pmatrix} 2 \\ - 3 \\ 1 \end{pmatrix}$ and passes through $A = (3,2,0)$ . The plane also contains $P = (1, - 3, - 1)$ , so

\overrightarrow{AP} = P - A = \begin{pmatrix} - 2 \\ - 5 \\ - 1 \end{pmatrix}

A normal to the plane is $\mathbf{d}_{1} \times \overrightarrow{AP}$ :

\mathbf{d}_{1} \times \overrightarrow{AP} = \begin{pmatrix} ( - 3)( - 1) - (1)( - 5) \\ (1)( - 2) - (2)( - 1) \\ (2)( - 5) - ( - 3)( - 2) \end{pmatrix} = \begin{pmatrix} 8 \\ 0 \\ - 16 \end{pmatrix}

Dividing by $8$ , a simpler normal vector is $\begin{pmatrix} 1 \\ 0 \\ - 2 \end{pmatrix}$ . The cartesian equation of the plane is $x - 2z = d$ , and substituting $A = (3,2,0)$ :

d = 3 - 2(0) = 3

To check your answer: substituting

P = (1, - 3, - 1)

:

1 - 2( - 1) = 3

, matching the plane’s constant. ✓

$\therefore$ the plane has cartesian equation $x - 2z = 3$ .

Part (b)

2 marks

$l_{1}$ has direction $\mathbf{d}_{1} = \begin{pmatrix} 2 \\ - 3 \\ 1 \end{pmatrix}$ and $l_{2}$ has direction $\mathbf{d}_{2} = \begin{pmatrix} 1 \\ 2 \\ 4 \end{pmatrix}$ .

\mathbf{d}_{1} \cdot \mathbf{d}_{2} = (2)(1) + ( - 3)(2) + (1)(4) = 2 - 6 + 4 = 0

Since the dot product of their direction vectors is $0$ , $l_{1}\bot l_{2}$ (shown).

Part (c)(i)

6 marks

Approach

the common perpendicular between two skew lines meets each at a specific parameter value. We find $\mathbf{r}_{1} - \mathbf{r}_{2}$ in terms of $\lambda$ and $\mu$ , then impose perpendicularity to both direction vectors as two simultaneous equations.

\mathbf{r}_{1} - \mathbf{r}_{2} = \begin{pmatrix} 3 \\ 2 \\ 0 \end{pmatrix} + \lambda\begin{pmatrix} 2 \\ - 3 \\ 1 \end{pmatrix} - \begin{pmatrix} - 4 \\ 1 \\ - 3 \end{pmatrix} - \mu\begin{pmatrix} 1 \\ 2 \\ 4 \end{pmatrix} = \begin{pmatrix} 7 + 2\lambda - \mu \\ 1 - 3\lambda - 2\mu \\ 3 + \lambda - 4\mu \end{pmatrix}

For $\mathbf{r}_{1} - \mathbf{r}_{2}$ to be perpendicular to $\mathbf{d}_{1}$ :

2(7 + 2\lambda - \mu) - 3(1 - 3\lambda - 2\mu) + 1(3 + \lambda - 4\mu) = 0

\Rightarrow 14 + 14\lambda + 0\mu = 0

\Rightarrow \lambda = - 1

For $\mathbf{r}_{1} - \mathbf{r}_{2}$ to be perpendicular to $\mathbf{d}_{2}$ :

1(7 + 2\lambda - \mu) + 2(1 - 3\lambda - 2\mu) + 4(3 + \lambda - 4\mu) = 0

\Rightarrow 21 + 0\lambda - 21\mu = 0

\Rightarrow \mu = 1

To check your answer: with

\lambda = - 1,\ \mu = 1

:

\mathbf{r}_{1} - \mathbf{r}_{2} = \begin{pmatrix} 4 \\ 2 \\ - 2 \end{pmatrix}

, and

\begin{pmatrix} 4 \\ 2 \\ - 2 \end{pmatrix} \cdot \mathbf{d}_{1} = 8 - 6 - 2 = 0

,

\begin{pmatrix} 4 \\ 2 \\ - 2 \end{pmatrix} \cdot \mathbf{d}_{2} = 4 + 4 - 8 = 0

. ✓

Substituting back into the line equations, the common perpendicular meets $l_{1}$ at

\begin{pmatrix} 3 \\ 2 \\ 0 \end{pmatrix} + ( - 1)\begin{pmatrix} 2 \\ - 3 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ 5 \\ - 1 \end{pmatrix}

and meets $l_{2}$ at

\begin{pmatrix} - 4 \\ 1 \\ - 3 \end{pmatrix} + (1)\begin{pmatrix} 1 \\ 2 \\ 4 \end{pmatrix} = \begin{pmatrix} - 3 \\ 3 \\ 1 \end{pmatrix}

$\therefore$ $\lambda = - 1$ , $\mu = 1$ ; the common perpendicular meets $l_{1}$ at $\begin{pmatrix} 1 \\ 5 \\ - 1 \end{pmatrix}$ and $l_{2}$ at $\begin{pmatrix} - 3 \\ 3 \\ 1 \end{pmatrix}$ .

Part (c)(ii)

2 marks

The length of the common perpendicular is the distance between the two meeting points found in (i).

\left| \begin{pmatrix} 1 \\ 5 \\ - 1 \end{pmatrix} - \begin{pmatrix} - 3 \\ 3 \\ 1 \end{pmatrix} \right|

= \left| \begin{pmatrix} 4 \\ 2 \\ - 2 \end{pmatrix} \right|

= \sqrt{4^{2} + 2^{2} + ( - 2)^{2}}

= \sqrt{24}

= 2\sqrt{6}

To check your answer:

2\sqrt{6} \approx 4.90

; since

\begin{pmatrix} 4 \\ 2 \\ - 2 \end{pmatrix}

is genuinely perpendicular to both line directions (confirmed in (i)), this is indeed the shortest distance between the two skew lines. ✓

$\therefore$ the length of the common perpendicular is $2\sqrt{6}$ units.

2021 H2 Maths Paper 1, Question 8

Part (a)

Part (b)

Part (c)(i)

Part (c)(ii)

This is how we mark, and how we teach.