2022 H2 Maths Paper 1 Q11 — Worked Solution

Part (a)

3 marks

Let $P = (1136,92,p)$ and $Q = (200,20, - 15)$ .

\overrightarrow{QP} = P - Q = \left( 1136 - 200,\ 92 - 20,\ p - ( - 15) \right) = (936,\ 72,\ p + 15)

Since $PQ = 939$ ,

936^{2} + 72^{2} + (p + 15)^{2} = 939^{2}

\Rightarrow 876096 + 5184 + (p + 15)^{2} = 881721

\Rightarrow (p + 15)^{2} = 441

\Rightarrow p + 15 = \pm 21

\Rightarrow p = 6\ \text{or}\ p = - 36

Since $P$ is below $Q$ in level (the $z$ -coordinate), we need $p < - 15$ . Only $p = - 36$ satisfies this $\left( p = 6\ \text{rej}\ \text{since}\ 6 > - 15 \right)$ .

To check your answer: with

p = - 36

:

\overrightarrow{QP} = (936,72, - 21)

, so

PQ = \sqrt{936^{2} + 72^{2} + 21^{2}} = \sqrt{881721} = 939

✓

$\therefore$ $p = - 36$ .

Part (b)

4 marks

Let the three given points be $A(400,600, - 20)$ , $B(500,200, - 70)$ , $C(600, - 340, - 50)$ .

\overrightarrow{AB} = B - A = (100, - 400, - 50),\quad\overrightarrow{AC} = C - A = (200, - 940, - 30)

A normal to the plane is $\overrightarrow{AB} \times \overrightarrow{AC}$ :

\overrightarrow{AB} \times \overrightarrow{AC} = \begin{pmatrix} ( - 400)( - 30) - ( - 50)( - 940) \\ ( - 50)(200) - (100)( - 30) \\ (100)( - 940) - ( - 400)(200) \end{pmatrix} = \begin{pmatrix} - 35000 \\ - 7000 \\ - 14000 \end{pmatrix}

Dividing by $- 7000$ , a simpler normal vector is $(5,1,2)$ . The cartesian equation of the plane is $5x + y + 2z = d$ , and substituting $A(400,600, - 20)$ :

d = 5(400) + 600 + 2( - 20) = 2560

To check your answer: substituting

B(500,200, - 70)

:

5(500) + 200 + 2( - 70) = 2560

✓; substituting

C(600, - 340, - 50)

:

5(600) - 340 + 2( - 50) = 2560

✓

$\therefore$ $5x + y + 2z = 2560$ .

Part (c)

4 marks

Approach

parametrise the line $PQ$ , substitute into the plane equation to find the parameter at the point of intersection, and check it lies between $P$ and $Q$ .

The pipeline $PQ$ has direction $\overrightarrow{PQ} = Q - P = ( - 936, - 72,21)$ , so a vector equation of the line through $P$ and $Q$ is

\mathbf{r} = \begin{pmatrix} 1136 \\ 92 \\ - 36 \end{pmatrix} + t\begin{pmatrix} - 936 \\ - 72 \\ 21 \end{pmatrix},\quad t \in \mathbb{R}

Substituting into $5x + y + 2z = 2560$ :

5(1136 - 936t) + (92 - 72t) + 2( - 36 + 21t) = 2560

\Rightarrow 5700 - 4710t = 2560

\Rightarrow t = \frac{3140}{4710} = \frac{2}{3}

Since $0 < \frac{2}{3} < 1$ , this point lies between $P$ and $Q$ , so the pipeline does pass through the rock layer. Substituting $t = \frac{2}{3}$ :

x = 1136 - 936\left( \frac{2}{3} \right) = 512,\quad y = 92 - 72\left( \frac{2}{3} \right) = 44,\quad z = - 36 + 21\left( \frac{2}{3} \right) = - 22

To check your answer:

5(512) + 44 + 2( - 22) = 2560 + 44 - 44 = 2560

✓, so

(512,44, - 22)

lies on the plane

$\therefore$ the pipeline meets the rock layer at $(512,44, - 22)$ .

Part (d)

2 marks

The horizontal plane has normal $\mathbf{k} = (0,0,1)$ . The direction of the pipeline is $\mathbf{d} = (936,72, - 21)$ , with $\left| \mathbf{d} \right| = 939$ (from part (a)). If $\theta$ is the angle between the pipeline and the horizontal, $\theta$ is the complement of the angle between $\mathbf{d}$ and $\mathbf{k}$ , so

\sin\theta = \frac{\left| \mathbf{d} \cdot \mathbf{k} \right|}{\left| \mathbf{d} \right|\left| \mathbf{k} \right|} = \frac{| - 21|}{939} = \frac{7}{313}

\Rightarrow \theta = \sin^{- 1}\left( \frac{7}{313} \right) = {1.28}^{\circ}\ \left( \text{3 s.f.} \right)

To check your answer:

\frac{7}{313} \approx 0.0224

, and

\sin^{- 1}(0.0224) \approx {1.28}^{\circ}

; this is a small angle, consistent with the pipeline’s

z

-displacement (

21\, m

) being much smaller than its horizontal displacement (

\text{≈}938.8\, m

) ✓

$\therefore$ $\theta = {1.28}^{\circ}$ .

2022 H2 Maths Paper 1, Question 11

Part (a)

Part (b)

Part (c)

Part (d)

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