8.7.5 Vectors, SPM Practice (Question 9) - SPM Additional Mathematics

Question 9:

Diagram 2 shows a triangle EFG.

It is given E R : R F = 1 : 2, F G : T G = 3 : 1, \vec{E R} = 4 \underset{˜}{x} and \vec{E G} = 6 \underset{˜}{y} .

\begin{array}{l} a) Express in terms of \underset{˜}{x} and \underset{˜}{y} : \\ (i) \vec{G R} \\ (ii) \vec{G T} \end{array}

[3 marks]
(b) If line GR is extended to point K such that

\vec{G K} = h \vec{G R} and \vec{E K} = 6 \underset{˜}{x} - 3 \underset{˜}{y}, find the value of h .

[3 marks]

Solution:

(a)(i)

\begin{array}{l} \vec{G R} = \vec{G E} + \vec{E R} \\ = - 6 \underset{˜}{y} + 4 \underset{˜}{x} \end{array}

(a)(ii)

\begin{array}{l} \vec{G T} = \frac{1}{3} \vec{G F} \\ = \frac{1}{3} (\vec{G R} + \vec{R F}) \\ = \frac{1}{3} (- 6 \underset{˜}{y} + 4 \underset{˜}{x} + 8 \underset{˜}{x}) \leftarrow \vec{R F} = 2 \vec{E R} \\ = - 2 \underset{˜}{y} + 4 \underset{˜}{x} \end{array}

(b)

\begin{array}{l} \vec{G K} = h \vec{G R} \\ \vec{G E} + \vec{E K} = h \vec{G R} \\ - 6 \underset{˜}{y} + (6 \underset{˜}{x} - 3 \underset{˜}{y}) = h (- 6 \underset{˜}{y} + 4 \underset{˜}{x}) \\ - 9 \underset{˜}{y} + 6 \underset{˜}{x} = - 6 h \underset{˜}{y} + 4 h \underset{˜}{x} \\ Compare, \\ - 6 h = - 9 \\ h = \frac{3}{2} \end{array}