8.6.2 Vector Short Questions (Question 4 & 5)

Question 4:

Diagram below shows a parallelogram ABCD with BED as a straight line.

\begin{matrix} Given that - - \to A B = 7 p ˜, - - \to A D = 5 q ˜ and D E = 3 E B, express, in terms of p ˜ and q ˜ . (a) - - \to B D (b) - - \to E C \end{matrix}

$\begin{array}{l} Given that \vec{A B} = 7 \underset{˜}{p}, \vec{A D} = 5 \underset{˜}{q} and D E = 3 E B, express, in terms of \underset{˜}{p} and \underset{˜}{q} . \\ (a) \vec{B D} \\ (b) \vec{E C} \end{array}$

Solution:
(a)

\begin{matrix} Note: for parallelogram, - - \to A B = - - \to D C = 7 p ˜, - - \to A D = - - \to B C = 5 q ˜ . - - \to B D = - - \to B A + - - \to A D - - \to B D = - 7 p ˜ + 5 q ˜ \end{matrix}

$\begin{array}{l} Note: for parallelogram, \vec{A B} = \vec{D C} = 7 \underset{˜}{p}, \vec{A D} = \vec{B C} = 5 \underset{˜}{q} . \\ \vec{B D} = \vec{B A} + \vec{A D} \\ \vec{B D} = - 7 \underset{˜}{p} + 5 \underset{˜}{q} \end{array}$

(b)

\begin{matrix} - - \to D E = 3 - - \to E B \frac{- - \to E B}{- - \to D E} = \frac{1}{3} \to E B : D E = 1 : 3 ∴ - - \to E B = \frac{1}{4} - - \to D B = \frac{1}{4} (- - - \to B D) = \frac{1}{4} [- (- 7 p ˜ + 5 q ˜)] \leftarrow F r o m (a) = \frac{7}{4} p ˜ - \frac{5}{4} q ˜ \end{matrix}

$\begin{array}{l} \vec{D E} = 3 \vec{E B} \\ \frac{\vec{E B}}{\vec{D E}} = \frac{1}{3} \to E B : D E = 1 : 3 \\ ∴ \vec{E B} = \frac{1}{4} \vec{D B} \\ = \frac{1}{4} (- \vec{B D}) \\ = \frac{1}{4} [- (- 7 \underset{˜}{p} + 5 \underset{˜}{q})] \leftarrow F r o m (a) \\ = \frac{7}{4} \underset{˜}{p} - \frac{5}{4} \underset{˜}{q} \end{array}$

\begin{matrix} - - \to E C = - - \to E B + - - \to B C - - \to E C = \frac{7}{4} p ˜ - \frac{5}{4} q ˜ + 5 q ˜ - - \to E C = \frac{7}{4} p ˜ + \frac{15}{4} q ˜ \end{matrix}

$\begin{array}{l} \vec{E C} = \vec{E B} + \vec{B C} \\ \vec{E C} = \frac{7}{4} \underset{˜}{p} - \frac{5}{4} \underset{˜}{q} + 5 \underset{˜}{q} \\ \vec{E C} = \frac{7}{4} \underset{˜}{p} + \frac{15}{4} \underset{˜}{q} \end{array}$

Question 5:

Use the above information to find the values of h and k when r = 2p – 3q.

Solution:

\begin{matrix} r = 2 p - 3 q (h - 1) a ˜ + (h + k) b ˜ = 2 (5 a ˜ - 7 b ˜) - 3 (- 2 a ˜ + 3 b ˜) (h - 1) a ˜ + (h + k) b ˜ = 10 a ˜ - 14 b ˜ + 6 a ˜ - 9 b ˜ (h - 1) a ˜ + (h + k) b ˜ = 16 a ˜ - 23 b ˜ Comparing vector: h - 1 = 16 h = 17 h + k = - 23 17 + k = - 23 k = - 40 \end{matrix}

$\begin{array}{l} r = 2 p - 3 q \\ (h - 1) \underset{˜}{a} + (h + k) \underset{˜}{b} = 2 (5 \underset{˜}{a} - 7 \underset{˜}{b}) - 3 (- 2 \underset{˜}{a} + 3 \underset{˜}{b}) \\ (h - 1) \underset{˜}{a} + (h + k) \underset{˜}{b} = 10 \underset{˜}{a} - 14 \underset{˜}{b} + 6 \underset{˜}{a} - 9 \underset{˜}{b} \\ (h - 1) \underset{˜}{a} + (h + k) \underset{˜}{b} = 16 \underset{˜}{a} - 23 \underset{˜}{b} \\ Comparing vector: \\ h - 1 = 16 \\ h = 17 \\ h + k = - 23 \\ 17 + k = - 23 \\ k = - 40 \end{array}$

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