2015 H2 Maths Paper 2 Q2 — Worked Solution

Part (i)

2 marks

Approach

the angle between a line and a coordinate axis is found from the direction vector of the line and the unit vector along that axis, using the acute-angle (absolute-value) form of the angle formula.

$L$ has direction vector $\mathbf{d} = 2\mathbf{i} + 3\mathbf{j} - 6\mathbf{k}$ , and the $x$ -axis has direction $\mathbf{i}$ .

\left| \mathbf{d} \right| = \sqrt{2^{2} + 3^{2} + ( - 6)^{2}} = \sqrt{49} = 7

Let $\theta$ be the acute angle between $L$ and the $x$ -axis. Then

\cos\theta = \frac{\left| \mathbf{d} \cdot \mathbf{i} \right|}{\left| \mathbf{d} \right|\left| \mathbf{i} \right|}

= \frac{2}{7}

By GC, $\theta = 73.4{^\circ}$ (1 d.p.).

Part (ii)

5 marks

Approach

a general point on $L$ is $Q = (1 + 2\lambda,\ - 2 + 3\lambda,\ - 4 - 6\lambda)$ . We set $\left| \overrightarrow{PQ} \right|^{2} = 33$ and solve for $\lambda$ . Since $\left| \overrightarrow{PQ} \right|^{2}$ is quadratic in $\lambda$ , the closest point (its minimum) occurs at the value of $\lambda$ midway between the two roots, by symmetry of the parabola about its vertex — this is quicker than differentiating afresh.

Since $P = 2\mathbf{i} + 5\mathbf{j} - 6\mathbf{k}$ ,

\overrightarrow{PQ} = (2\lambda - 1)\mathbf{i} + (3\lambda - 7)\mathbf{j} + (2 - 6\lambda)\mathbf{k}

\left| \overrightarrow{PQ} \right|^{2} = (2\lambda - 1)^{2} + (3\lambda - 7)^{2} + (2 - 6\lambda)^{2}

= 49\lambda^{2} - 70\lambda + 54

Setting this equal to $33$ :

49\lambda^{2} - 70\lambda + 54 = 33

49\lambda^{2} - 70\lambda + 21 = 0

7\lambda^{2} - 10\lambda + 3 = 0

(7\lambda - 3)(\lambda - 1) = 0

\lambda = \frac{3}{7}\ \ \text{or}\ \ \lambda = 1

Substituting back into the coordinates of $Q$ :

\lambda = \frac{3}{7}:\ \ Q = \left( \frac{13}{7},\ - \frac{5}{7},\ - \frac{46}{7} \right)

\lambda = 1:\ \ Q = (3,\ 1,\ - 10)

$\therefore$ the points on $L$ at distance $\sqrt{33}$ from $P$ are $\left( \frac{13}{7},\ - \frac{5}{7},\ - \frac{46}{7} \right)$ and $(3,\ 1,\ - 10)$ .

Hence, the point on $L$ closest to $P$ occurs at the value of $\lambda$ midway between these two roots:

\lambda = \frac{1}{2}\left( \frac{3}{7} + 1 \right) = \frac{5}{7}

Q_{\min} = \left( 1 + 2 \times \frac{5}{7},\ - 2 + 3 \times \frac{5}{7},\ - 4 - 6 \times \frac{5}{7} \right) = \left( \frac{17}{7},\ \frac{1}{7},\ - \frac{58}{7} \right)

To check your answer: at

\lambda = \frac{5}{7}

,

\left| \overrightarrow{PQ} \right|^{2} = 49\left( \frac{5}{7} \right)^{2} - 70\left( \frac{5}{7} \right) + 54 = 29

, and

29 < 33

, consistent with this being the closest point on

L

to

P

. ✓

$\therefore$ the point on $L$ closest to $P$ is $\left( \frac{17}{7},\ \frac{1}{7},\ - \frac{58}{7} \right)$ .

Part (iii)

3 marks

Approach

a plane containing $L$ and $P$ is fixed by two directions lying in it: the direction of $L$ , and the direction from a point on $L$ to $P$ . Their cross product gives a normal to the plane.

Let $A = (1,\ - 2,\ - 4)$ , a point on $L$ . Then

\overrightarrow{AP} = (2 - 1)\mathbf{i} + \left( 5 - ( - 2) \right)\mathbf{j} + \left( - 6 - ( - 4) \right)\mathbf{k} = \mathbf{i} + 7\mathbf{j} - 2\mathbf{k}

A normal to the plane is

\mathbf{n} = \mathbf{d} \times \overrightarrow{AP}

= \left( 3 \times ( - 2) - ( - 6) \times 7 \right)\mathbf{i} + \left( ( - 6) \times 1 - 2 \times ( - 2) \right)\mathbf{j} + (2 \times 7 - 3 \times 1)\mathbf{k}

= 36\mathbf{i} - 2\mathbf{j} + 11\mathbf{k}

Using $A$ as a point on the plane,

\mathbf{n} \cdot \overrightarrow{OA} = 36(1) - 2( - 2) + 11( - 4) = - 4

To check your answer: substituting

P = (2,\ 5,\ - 6)

:

36(2) - 2(5) + 11( - 6) = 72 - 10 - 66 = - 4

, matching the plane’s constant, confirming

P

lies in the plane. ✓

$\therefore$ a cartesian equation of the plane is $36x - 2y + 11z = - 4$ .

2015 H2 Maths Paper 2, Question 2

Part (i)

Part (ii)

Part (iii)

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