2023 H2 Maths Paper 1 Q9 — Worked Solution

Part (a)

5 marks

Since $l_{1}$ passes through $A( - 3,1,2)$ with direction $\begin{pmatrix} 2 \\ 1 \\ a \end{pmatrix}$ ,

l_{1}:\ \mathbf{r} = \begin{pmatrix} - 3 \\ 1 \\ 2 \end{pmatrix} + s\begin{pmatrix} 2 \\ 1 \\ a \end{pmatrix}

Writing $l_{2}$ in vector form, letting $\frac{x + 2}{3} = \frac{y - 1}{2} = z - 5 = t$ ,

l_{2}:\ \mathbf{r} = \begin{pmatrix} - 2 \\ 1 \\ 5 \end{pmatrix} + t\begin{pmatrix} 3 \\ 2 \\ 1 \end{pmatrix}

Since $l_{1}$ and $l_{2}$ cross at $B$ , equating components:

- 3 + 2s = - 2 + 3t\quad...(1)

1 + s = 1 + 2t\quad...(2)

2 + as = 5 + t\quad...(3)

Equations (1) and (2) involve only $s$ and $t$ , so we solve these first. From (2), $s = 2t$ . Substituting into (1):

- 3 + 2(2t) = - 2 + 3t

\Rightarrow - 3 + 4t = - 2 + 3t

\Rightarrow t = 1

So $s = 2t = 2$ .

Substituting $s = 2$ , $t = 1$ into (3):

2 + 2a = 5 + 1

\Rightarrow 2a = 4

\Rightarrow a = 2

From $l_{2}$ with $t = 1$ , $B = ( - 2 + 3,\ 1 + 2,\ 5 + 1) = (1,3,6)$ .

To check your answer: substituting

s = 2

into

l_{1}

gives

x = - 3 + 4 = 1

,

y = 1 + 2 = 3

(independent of

a

), matching

B

found from

l_{2}

. ✓

$\therefore$ $a = 2$ and $B = (1,3,6)$ .

Part (b)(i)

3 marks

With $a = 2$ , $l_{1}$ has direction $\mathbf{d}_{1} = \begin{pmatrix} 2 \\ 1 \\ 2 \end{pmatrix}$ . Since $\pi_{1}$ is perpendicular to $l_{2}$ , the normal of $\pi_{1}$ is the direction vector of $l_{2}$ , $\mathbf{n}_{1} = \begin{pmatrix} 3 \\ 2 \\ 1 \end{pmatrix}$ . Since $\pi_{1}$ contains $A( - 3,1,2)$ ,

\mathbf{n}_{1} \cdot \left( \mathbf{r} - A \right) = 0

\Rightarrow 3(x + 3) + 2(y - 1) + (z - 2) = 0

\Rightarrow \pi_{1}:\ 3x + 2y + z = - 5

The shortest (perpendicular) distance from $B(1,3,6)$ to $\pi_{1}$ is

d = \frac{\left| 3(1) + 2(3) + 6 - ( - 5) \right|}{\sqrt{3^{2} + 2^{2} + 1^{2}}}

= \frac{|3 + 6 + 6 + 5|}{\sqrt{14}}

= \frac{20}{\sqrt{14}}

= \frac{10\sqrt{14}}{7}

\approx 5.35

$\therefore$ the shortest distance from $B$ to $\pi_{1}$ is $\frac{10\sqrt{14}}{7} \approx 5.35$ (3 s.f.).

Part (b)(ii)

2 marks

Let $\theta$ be the acute angle between $l_{1}$ and $\pi_{1}$ (measured from the line to the plane, not to the normal):

\sin\theta = \frac{\left| \mathbf{d}_{1} \cdot \mathbf{n}_{1} \right|}{\left| \mathbf{d}_{1} \right|\left| \mathbf{n}_{1} \right|}

\mathbf{d}_{1} \cdot \mathbf{n}_{1} = (2)(3) + (1)(2) + (2)(1) = 10

\left| \mathbf{d}_{1} \right| = \sqrt{4 + 1 + 4} = 3,\quad\left| \mathbf{n}_{1} \right| = \sqrt{9 + 4 + 1} = \sqrt{14}

\sin\theta = \frac{10}{3\sqrt{14}}

\Rightarrow \theta = \sin^{- 1}\left( \frac{10}{3\sqrt{14}} \right)

\approx {63.0}^{\circ}\ \left( 1\text{ d.p.} \right)

$\therefore$ the acute angle between $l_{1}$ and $\pi_{1}$ is ${63.0}^{\circ}$ .

Part (c)

3 marks

Approach: $\pi_{2}$ contains $l_{1}$ , so its normal $\mathbf{n}_{2}$ is perpendicular to $\mathbf{d}_{1}$ ; $\pi_{2}$ is also perpendicular to $\pi_{1}$ , so $\mathbf{n}_{2}$ is perpendicular to $\mathbf{n}_{1}$ too. A vector perpendicular to both $\mathbf{n}_{1}$ and $\mathbf{d}_{1}$ is $\mathbf{n}_{1} \times \mathbf{d}_{1}$ .

\mathbf{n}_{2} = \mathbf{n}_{1} \times \mathbf{d}_{1} = \begin{pmatrix} 3 \\ 2 \\ 1 \end{pmatrix} \times \begin{pmatrix} 2 \\ 1 \\ 2 \end{pmatrix}

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

= \begin{pmatrix} 3 \\ - 4 \\ - 1 \end{pmatrix}

To check your answer:

\mathbf{n}_{2} \cdot \mathbf{n}_{1} = 9 - 8 - 1 = 0

and

\mathbf{n}_{2} \cdot \mathbf{d}_{1} = 6 - 4 - 2 = 0

, so

\mathbf{n}_{2}

is perpendicular to both, as required. ✓

Since $\pi_{2}$ contains $A( - 3,1,2)$ ,

\mathbf{n}_{2} \cdot \left( \mathbf{r} - A \right) = 0

\Rightarrow 3(x + 3) - 4(y - 1) - (z - 2) = 0

\Rightarrow \pi_{2}:\ 3x - 4y - z = - 15

$\therefore$ $\pi_{2}:3x - 4y - z = - 15$ .

2023 H2 Maths Paper 1, Question 9

Part (a)

Part (b)(i)

Part (b)(ii)

Part (c)

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