SPM Additional Mathematics 2019, Paper 2 (Question 6)

Question 6:
Solution by scale drawing is not accepted.
Diagram 3 shows the positions of jetty O and kelongs K, L, R, S and T in the sea.

Diagram 3

Kelong L is situated 400 m from jetty O and kelong R is situated 600 m from jetty O in the direction of OL. Kelong S is situated 300 m from jetty O and kelong T is situated 600 m from kelong 5 in the directions of OS. Kelongs L, K and T are situated on a straight line such that the distance of kelong K from kelong T is 5 times its distance from kelong L.

(a) By using to represent 100 m in the direction of OR and to represent 150 m in the direction of OT, express in terms of

\underset{˜}{p} and \underset{˜}{q} .

$\underset{˜}{p} and \underset{˜}{q} .$
[3 marks]

(i) \vec{O K}, (ii) \vec{R K} .

$(i) \vec{O K}, (ii) \vec{R K} .$

(b) If Joe uses a binocular to observe kelong R from kelong S, determine whether kelong R can be seen without being blocked by kelong K or otherwise.
Prove you answer mathematically. [5 marks]

Solution:
(a)

\begin{array}{l} K T = 5 K L \\ \frac{K T}{K L} = \frac{5}{1} \\ K T : K L = 5 : 1 \\ \vec{O L} = 4 \underset{˜}{p}, \vec{L R} = 2 \underset{˜}{p}, \vec{O R} = 6 \underset{˜}{p} \\ \vec{O S} = 2 \underset{˜}{q}, \vec{S T} = 4 \underset{˜}{q}, \vec{O T} = 6 \underset{˜}{q} \end{array}

$\begin{array}{l} K T = 5 K L \\ \frac{K T}{K L} = \frac{5}{1} \\ K T : K L = 5 : 1 \\ \vec{O L} = 4 \underset{˜}{p}, \vec{L R} = 2 \underset{˜}{p}, \vec{O R} = 6 \underset{˜}{p} \\ \vec{O S} = 2 \underset{˜}{q}, \vec{S T} = 4 \underset{˜}{q}, \vec{O T} = 6 \underset{˜}{q} \end{array}$

(a)(i)

\begin{array}{l} \vec{O K} = \vec{O L} + \vec{L K} \\ = 4 \underset{˜}{p} + \frac{1}{6} \vec{L T} \\ = 4 \underset{˜}{p} + \frac{1}{6} (- 4 \underset{˜}{p} + 6 \underset{˜}{q}) \\ = 4 \underset{˜}{p} - \frac{2}{3} \underset{˜}{p} + \underset{˜}{q} \\ = \frac{10}{3} \underset{˜}{p} + \underset{˜}{q} \end{array}

$\begin{array}{l} \vec{O K} = \vec{O L} + \vec{L K} \\ = 4 \underset{˜}{p} + \frac{1}{6} \vec{L T} \\ = 4 \underset{˜}{p} + \frac{1}{6} (- 4 \underset{˜}{p} + 6 \underset{˜}{q}) \\ = 4 \underset{˜}{p} - \frac{2}{3} \underset{˜}{p} + \underset{˜}{q} \\ = \frac{10}{3} \underset{˜}{p} + \underset{˜}{q} \end{array}$

(a)(ii)

\begin{array}{l} \vec{R K} = \vec{R O} + \vec{O K} \\ = - 6 \underset{˜}{p} + \frac{10}{3} \underset{˜}{p} + \underset{˜}{q} \\ = - \frac{8}{3} \underset{˜}{p} + \underset{˜}{q} \end{array}

$\begin{array}{l} \vec{R K} = \vec{R O} + \vec{O K} \\ = - 6 \underset{˜}{p} + \frac{10}{3} \underset{˜}{p} + \underset{˜}{q} \\ = - \frac{8}{3} \underset{˜}{p} + \underset{˜}{q} \end{array}$

(b)

\begin{array}{l} \vec{S R} = \vec{O R} - \vec{O S} \\ = 6 \underset{˜}{p} - 2 \underset{˜}{q} \\ = 2 (3 \underset{˜}{p} - \underset{˜}{q}) \\ \vec{K R} = - \vec{R K} \\ = - (- \frac{8}{3} \underset{˜}{p} + \underset{˜}{q}) \\ = \frac{8}{3} \underset{˜}{p} - \underset{˜}{q} \\ \vec{S R} \neq m \vec{K R}, m is a constant . \end{array}

$\begin{array}{l} \vec{S R} = \vec{O R} - \vec{O S} \\ = 6 \underset{˜}{p} - 2 \underset{˜}{q} \\ = 2 (3 \underset{˜}{p} - \underset{˜}{q}) \\ \vec{K R} = - \vec{R K} \\ = - (- \frac{8}{3} \underset{˜}{p} + \underset{˜}{q}) \\ = \frac{8}{3} \underset{˜}{p} - \underset{˜}{q} \\ \vec{S R} \neq m \vec{K R}, m is a constant . \end{array}$
S, K and R are not collinear.
Hence, Kelong R can be seen without being blocked by Kelong K.

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