2019 H2 Maths Paper 1 Q12 — Worked Solution

Part (i)

5 marks

Approach

$Q$ is the point where the line through $P$ in the direction of the ray meets the top plane; $R$ is the point where the line through $S$ , parallel to that same direction (since $PQ$ is parallel to $RS$ ), meets the base plane. Both are found by substituting a general point of the line into the plane equation and solving for the parameter.

The line along $PQ$ passes through $P(2,2,4)$ in direction $\begin{pmatrix} - 2 \\ - 3 \\ - 6 \end{pmatrix}$ , so a general point on it is

\begin{pmatrix} 2 - 2\lambda \\ 2 - 3\lambda \\ 4 - 6\lambda \end{pmatrix}.

Since $Q$ lies on the plane $x + y + z = 1$ ,

(2 - 2\lambda) + (2 - 3\lambda) + (4 - 6\lambda) = 1

8 - 11\lambda = 1

\lambda = \frac{7}{11}.

So

Q = \left( 2 - \frac{14}{11},\ 2 - \frac{21}{11},\ 4 - \frac{42}{11} \right) = \left( \frac{8}{11},\ \frac{1}{11},\ \frac{2}{11} \right).

Since $PQ$ is parallel to $RS$ , the line along $RS$ also has direction $\begin{pmatrix} - 2 \\ - 3 \\ - 6 \end{pmatrix}$ , so a general point of the line through $S( - 5, - 6, - 7)$ is

\begin{pmatrix} - 5 - 2\mu \\ - 6 - 3\mu \\ - 7 - 6\mu \end{pmatrix}.

Since $R$ lies on the plane $x + y + z = - 9$ ,

( - 5 - 2\mu) + ( - 6 - 3\mu) + ( - 7 - 6\mu) = - 9

- 18 - 11\mu = - 9

\mu = - \frac{9}{11}.

So

R = \left( - 5 + \frac{18}{11},\ - 6 + \frac{27}{11},\ - 7 + \frac{54}{11} \right) = \left( - \frac{37}{11},\ - \frac{39}{11},\ - \frac{23}{11} \right).

To check your answer: the coordinates of

R

sum to

- \frac{37}{11} - \frac{39}{11} - \frac{23}{11} = - 9

, matching the base plane equation. ✓

$\therefore$ $Q = \left( \frac{8}{11},\ \frac{1}{11},\ \frac{2}{11} \right)$ and $R = \left( - \frac{37}{11},\ - \frac{39}{11},\ - \frac{23}{11} \right)$ .

Part (ii)

3 marks

Let $\mathbf{n} = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$ be a normal to both the top and (parallel) base planes. Since $\theta$ and $\beta$ are the acute angles between the rays $PQ$ , $QR$ and this normal, we use the modulus of the dot product.

\overrightarrow{PQ} = Q - P = \begin{pmatrix} - 14/11 \\ - 21/11 \\ - 42/11 \end{pmatrix},\quad\quad\overrightarrow{QR} = R - Q = \begin{pmatrix} - 45/11 \\ - 40/11 \\ - 25/11 \end{pmatrix}.

\left| \overrightarrow{PQ} \right| = \frac{1}{11}\sqrt{14^{2} + 21^{2} + 42^{2}} = \frac{1}{11}\sqrt{2401} = \frac{49}{11},\quad\quad\overrightarrow{PQ} \cdot \mathbf{n} = - \frac{14 + 21 + 42}{11} = - 7.

Since $\theta$ is acute,

\cos\theta = \frac{\left| \overrightarrow{PQ} \cdot \mathbf{n} \right|}{\left| \overrightarrow{PQ} \right|\left| \mathbf{n} \right|}

= \frac{7}{(49/11)\sqrt{3}}

= \frac{77}{49\sqrt{3}}

= \frac{11}{7\sqrt{3}}

= \frac{11\sqrt{3}}{21}.

Similarly,

\left| \overrightarrow{QR} \right| = \frac{1}{11}\sqrt{45^{2} + 40^{2} + 25^{2}} = \frac{1}{11}\sqrt{4250} = \frac{5\sqrt{170}}{11},\quad\quad\overrightarrow{QR} \cdot \mathbf{n} = - \frac{45 + 40 + 25}{11} = - 10.

\cos\beta = \frac{\left| \overrightarrow{QR} \cdot \mathbf{n} \right|}{\left| \overrightarrow{QR} \right|\left| \mathbf{n} \right|}

= \frac{10}{\left( 5\sqrt{170}/11 \right)\sqrt{3}}

= \frac{10 \times 11}{5\sqrt{170}\sqrt{3}}

= \frac{22}{\sqrt{510}}

= \frac{11\sqrt{510}}{255}.

By GC, $\cos\theta = 0.907$ and $\cos\beta = 0.974$ (3 s.f.).

$\therefore$ $\cos\theta = \frac{11\sqrt{3}}{21} = 0.907$ and $\cos\beta = \frac{11\sqrt{510}}{255} = 0.974$ (3 s.f.).

Part (iii)

3 marks

Approach

The prism’s two faces are parallel planes; the thickness measured along the normal is simply the perpendicular distance between them, found from the plane constants divided by $\left| \mathbf{n} \right|$ .

The top and base planes, $x + y + z = 1$ and $x + y + z = - 9$ , are both normal to $\mathbf{n} = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$ , so the perpendicular distance between them is

\text{Thickness} = \frac{\left| 1 - ( - 9) \right|}{\left| \mathbf{n} \right|}

= \frac{10}{\sqrt{3}}

= \frac{10\sqrt{3}}{3}.

$\therefore$ the thickness of the prism measured in the direction of the normal at $Q$ is $\frac{10\sqrt{3}}{3} = 5.77$ (3 s.f.).

Part (iv)

1 mark

Using $\sin\theta = k\sin\beta$ with $\theta,\beta$ acute (so $\sin\theta,\sin\beta > 0$ ),

\sin\theta = \sqrt{1 - \cos^{2}\theta} = \sqrt{1 - \frac{121(3)}{441}} = \sqrt{\frac{78}{441}} = \frac{\sqrt{78}}{21},

\sin\beta = \sqrt{1 - \cos^{2}\beta} = \sqrt{1 - \frac{121(510)}{255^{2}}} = \sqrt{\frac{13}{255}}.

So

k = \frac{\sin\theta}{\sin\beta}

= \frac{\sqrt{78}/21}{\sqrt{13/255}}

= \frac{\sqrt{78}}{21}\sqrt{\frac{255}{13}}

= \frac{\sqrt{78 \times 255/13}}{21}

= \frac{\sqrt{1530}}{21}

= \frac{\sqrt{170}}{7},

using $78/13 = 6$ so $78 \times 255/13 = 6 \times 255 = 1530 = 9 \times 170$ , and $\sqrt{1530} = 3\sqrt{170}$ .

By GC, $k = 1.86$ (3 s.f.).

$\therefore$ $k = \frac{\sqrt{170}}{7} = 1.86$ (3 s.f.).

Part (v)

1 mark

Since $\sin\theta = k\sin\beta$ and $\theta,\beta \in \left( 0,\frac{\pi}{2} \right)$ , where $\sin$ is increasing, $\beta > \theta$ means $\sin\beta > \sin\theta$ . Then

k = \frac{\sin\theta}{\sin\beta} < 1.

$\therefore$ if $\beta > \theta$ , the refractive index $k < 1$ .

2019 H2 Maths Paper 1, Question 12

Part (i)

Part (ii)

Part (iii)

Part (iv)

Part (v)

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