8.7.3 Vectors, SPM Practice (Question 5 & 6)

Question 5:
Diagram below shows a triangle KLM.

$\begin{array}{l} It is given that K P : P L = 1 : 2, L R : R M = 2 : 1, \vec{K P} = 2 \underset{˜}{x}, \vec{K M} = 3 \underset{˜}{y} . \\ (a) Express in terms of \underset{˜}{x} and \underset{˜}{y}, \\ (i) \vec{M P} \\ (ii) \vec{M R} \\ (b) Given \underset{˜}{x} = 2 \underset{˜}{i} and \underset{˜}{y} = - \underset{˜}{i} + 4 \underset{˜}{j}, find \vec{| M R |} . \\ (c) Given \vec{M Q} = h \vec{M P} and \vec{Q R} = n \vec{K R}, where h and n are constants, \\ find the value of h and of n . \end{array}$

Solution:
(a)(i)

$\begin{array}{l} \vec{M P} = \vec{M K} + \vec{K P} \\ = - 3 \underset{˜}{y} + 2 \underset{˜}{x} \\ = 2 \underset{˜}{x} - 3 \underset{˜}{y} \end{array}$

(a)(ii)

$\begin{array}{l} \vec{M R} = \frac{1}{3} \vec{M L} \\ = \frac{1}{3} (\vec{M K} + \vec{K L}) \\ = \frac{1}{3} (- 3 \underset{˜}{y} + 6 \underset{˜}{x}) \\ = 2 \underset{˜}{x} - \underset{˜}{y} \end{array}$

(b)

$\begin{array}{l} \vec{M R} = 2 (2 \underset{˜}{i}) - (- \underset{˜}{i} + 4 \underset{˜}{j}) \\ = 4 \underset{˜}{i} + \underset{˜}{i} - 4 \underset{˜}{j} \\ = 5 \underset{˜}{i} - 4 \underset{˜}{j} \\ | \vec{M R} | = \sqrt{5^{2} + {(- 4)}^{2}} \\ = \sqrt{41} units \end{array}$

(c)

$\begin{array}{l} \vec{M Q} + \vec{Q R} = \vec{M R} \\ h \vec{M P} + n \vec{K R} = \vec{M R} \\ h (2 \underset{˜}{x} - 3 \underset{˜}{y}) + n (\vec{K M} + \vec{M R}) = 2 \underset{˜}{x} - \underset{˜}{y} \\ h (2 \underset{˜}{x} - 3 \underset{˜}{y}) + n (3 \underset{˜}{y} + 2 \underset{˜}{x} - \underset{˜}{y}) = 2 \underset{˜}{x} - \underset{˜}{y} \\ 2 h \underset{˜}{x} - 3 h \underset{˜}{y} + 2 n \underset{˜}{x} + 2 n \underset{˜}{y} = 2 \underset{˜}{x} - \underset{˜}{y} \\ (2 h + 2 n) \underset{˜}{x} + (- 3 h + 2 n) \underset{˜}{y} = 2 \underset{˜}{x} - \underset{˜}{y} \\ 2 h + 2 n = 2………. (1) \\ - 3 h + 2 n = - 1………. (2) \\ (1) - (2) : 5 h = 3 \\ h = \frac{3}{5} \\ From (1) : h + n = 1 \\ \frac{3}{5} + n = 1 \\ n = 1 - \frac{3}{5} \\ n = \frac{2}{5} \end{array}$

Question 6:
Diagram below shows a trapezium OABC and point D lies on AC.

$\begin{array}{l} It is given that \vec{O C} = 18 \underset{˜}{b}, \vec{O A} = 6 \underset{˜}{a} and \vec{O C} = 2 \vec{A B} . \\ (a) Express in terms of \underset{˜}{a} and \underset{˜}{b}, \\ (i) \vec{A C} \\ (ii) \vec{O B} \\ (b) It is given that \vec{A D} = k \vec{A C}, where k is a constant . \\ Find the value of k if the points O, D and B are collinear . \end{array}$

Solution:
(a)(i)

$\begin{array}{l} \vec{A C} = \vec{A O} + \vec{O C} \\ = - 6 \underset{˜}{a} + 18 \underset{˜}{b} \\ = 18 \underset{˜}{b} - 6 \underset{˜}{a} \end{array}$

(a)(ii)

$\begin{array}{l} \vec{O C} = 2 \vec{A B} \\ 18 \underset{˜}{b} = 2 (\vec{A O} + \vec{O B}) \\ 18 \underset{˜}{b} = 2 (- 6 \underset{˜}{a} + \vec{O B}) \\ 18 \underset{˜}{b} = - 12 \underset{˜}{a} + 2 \vec{O B} \\ \vec{O B} = 6 \underset{˜}{a} + 9 \underset{˜}{b} \end{array}$

(b)

$\begin{array}{l} \vec{O D} = h \vec{O B} \\ = h (6 \underset{˜}{a} + 9 \underset{˜}{b}) \\ = 6 h \underset{˜}{a} + 9 h \underset{˜}{b} \\ \vec{A D} = \vec{O D} - \vec{O A} \\ = 6 h \underset{˜}{a} + 9 h \underset{˜}{b} - 6 \underset{˜}{a} \\ = \underset{˜}{a} (6 h - 6) + 9 h \underset{˜}{b} \\ \vec{A D} = k \vec{A C} \\ \underset{˜}{a} (6 h - 6) + 9 h \underset{˜}{b} = k (18 \underset{˜}{b} - 6 \underset{˜}{a}) \\ \underset{˜}{a} (6 h - 6) + 9 h \underset{˜}{b} = - 6 k \underset{˜}{a} + 18 k \underset{˜}{b} \\ 6 h - 6 = - 6 k \\ h - 1 = - k \\ h = 1 - k ………. (1) \\ 9 h = 18 k \\ h = 2 k \\ From (1), \\ 1 - k = 2 k \\ 3 k = 1 \\ k = \frac{1}{3} \end{array}$

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