2017 H2 Maths Paper 1 Q10 — Worked Solution

Part (i)

4 marks

Approach

Both cables are modelled as lines through known points with known directions. If they meet, there is some point lying on both lines simultaneously, so we set up a common parameter for each line and solve for where the two position vectors agree.

Let $C$ have equation $\mathbf{r} = t\begin{pmatrix} 3 \\ 1 \\ - 2 \end{pmatrix}$ , $t \in \mathbb{R}$ , and let the new cable through $P(1,2, - 1)$ and $Q(5,7,a)$ have direction $\overrightarrow{PQ} = \begin{pmatrix} 4 \\ 5 \\ a + 1 \end{pmatrix}$ , so its equation is $\mathbf{r} = \begin{pmatrix} 1 \\ 2 \\ - 1 \end{pmatrix} + s\begin{pmatrix} 4 \\ 5 \\ a + 1 \end{pmatrix}$ , $s \in \mathbb{R}$ .

If the two cables meet, equating the $x$ , $y$ , $z$ components gives

3t = 1 + 4s

t = 2 + 5s

- 2t = - 1 + s(a + 1)

Substituting the second equation into the first:

3(2 + 5s) = 1 + 4s

6 + 15s = 1 + 4s

11s = - 5

s = - \frac{5}{11}

t = 2 + 5\left( - \frac{5}{11} \right) = - \frac{3}{11}

Substituting $t = - \frac{3}{11}$ and $s = - \frac{5}{11}$ into the third equation:

- 2\left( - \frac{3}{11} \right) = - 1 - \frac{5}{11}(a + 1)

\frac{6}{11} = - 1 - \frac{5}{11}(a + 1)

Multiplying throughout by $11$ :

6 = - 11 - 5(a + 1)

5(a + 1) = - 17

a + 1 = - \frac{17}{5}

a = - \frac{22}{5}

To check your answer: with

a = - \frac{22}{5}

, at

t = - \frac{3}{11}

the point on

C

is

\left( - \frac{9}{11}, - \frac{3}{11},\frac{6}{11} \right)

; at

s = - \frac{5}{11}

the point on the new cable is

\left( 1 - \frac{20}{11},2 - \frac{25}{11}, - 1 - \frac{5}{11}\left( \frac{3}{5} \right) \right) = \left( - \frac{9}{11}, - \frac{3}{11},\frac{6}{11} \right)

, which matches. ✓

$\therefore$ $a = - \frac{22}{5}$ .

Part (ii)

4 marks

Approach

Angle $\text{PRQ} = 90{^\circ}$ is equivalent to $\overrightarrow{RP} \cdot \overrightarrow{RQ} = 0$ . Writing $R$ as a general point on $C$ in terms of one parameter $t$ , this dot product becomes a quadratic in $t$ ; showing it has no real root shows no such $R$ exists.

With $a = - 3$ , $Q = (5,7, - 3)$ . Let $R = t\begin{pmatrix} 3 \\ 1 \\ - 2 \end{pmatrix} = (3t,t, - 2t)$ be a general point on $C$ .

\overrightarrow{RP} = \begin{pmatrix} 1 - 3t \\ 2 - t \\ - 1 + 2t \end{pmatrix},\quad\overrightarrow{RQ} = \begin{pmatrix} 5 - 3t \\ 7 - t \\ - 3 + 2t \end{pmatrix}

\overrightarrow{RP} \cdot \overrightarrow{RQ}

= (1 - 3t)(5 - 3t) + (2 - t)(7 - t) + ( - 1 + 2t)( - 3 + 2t)

= \left( 9t^{2} - 18t + 5 \right) + \left( t^{2} - 9t + 14 \right) + \left( 4t^{2} - 8t + 3 \right)

= 14t^{2} - 35t + 22

For angle $\text{PRQ} = 90{^\circ}$ , we need $\overrightarrow{RP} \cdot \overrightarrow{RQ} = 0$ , i.e.

14t^{2} - 35t + 22 = 0

The discriminant is

( - 35)^{2} - 4(14)(22) = 1225 - 1232 = - 7

Since the discriminant is negative, this quadratic has no real roots, so there is no real value of $t$ , and hence no point $R$ on $C$ , for which $\overrightarrow{RP} \cdot \overrightarrow{RQ} = 0$ .

$\therefore$ angle $\text{PRQ}$ cannot equal $90{^\circ}$ for any point $R$ on $C$ . (shown)

Part (iii)

5 marks

Approach

$PR$ is smallest when $R$ is the foot of the perpendicular from $P$ to $C$ . With $R = (3t,t, - 2t)$ , $PR^{2}$ is a quadratic in $t$ with positive leading coefficient, so its minimum is found by differentiating and setting the derivative to zero.

With $R = (3t,t, - 2t)$ as before,

PR^{2}

= (1 - 3t)^{2} + (2 - t)^{2} + ( - 1 + 2t)^{2}

= \left( 9t^{2} - 6t + 1 \right) + \left( t^{2} - 4t + 4 \right) + \left( 4t^{2} - 4t + 1 \right)

= 14t^{2} - 14t + 6

Differentiating with respect to $t$ and setting to zero:

\frac{d}{dt}\left( PR^{2} \right) = 28t - 14 = 0

t = \frac{1}{2}

Since the coefficient of $t^{2}$ is positive, this is a minimum.

R = \left( 3\left( \frac{1}{2} \right),\ \frac{1}{2},\ - 2\left( \frac{1}{2} \right) \right) = \left( \frac{3}{2},\ \frac{1}{2},\ - 1 \right)

PR^{2} = 14\left( \frac{1}{4} \right) - 14\left( \frac{1}{2} \right) + 6 = \frac{7}{2} - 7 + 6 = \frac{5}{2}

PR = \sqrt{\frac{5}{2}} = \frac{\sqrt{10}}{2}

To check your answer:

\overrightarrow{PR} = \left( \frac{3}{2} - 1,\ \frac{1}{2} - 2,\ - 1 - ( - 1) \right) = \left( \frac{1}{2},\ - \frac{3}{2},\ 0 \right)

, and this is perpendicular to the direction of

C

:

\left( \frac{1}{2} \right)(3) + \left( - \frac{3}{2} \right)(1) + 0( - 2) = \frac{3}{2} - \frac{3}{2} = 0

, which matches. ✓

$\therefore$ $R = \left( \frac{3}{2},\ \frac{1}{2},\ - 1 \right)$ , and the minimum length of $PR$ is $\frac{\sqrt{10}}{2}\ m$ .

2017 H2 Maths Paper 1, Question 10

Part (i)

Part (ii)

Part (iii)

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