8.7.1 Vectors, Long Questions (Question 1 & 2)

Question 1:

The above diagram shows triangle OAB. The straight line AP intersects the straight line OQ at R. It is given that

O P = \frac{1}{4} O B, A Q = \frac{1}{4} A B, - - \to O P = 4 b ˜ and - - \to O A = 8 a ˜ .

$O P = \frac{1}{4} O B, A Q = \frac{1}{4} A B, \vec{O P} = 4 \underset{˜}{b} and \vec{O A} = 8 \underset{˜}{a} .$

(a) Express in terms of

a ˜ and/ or b ˜ :

$\underset{˜}{a} and/ or \underset{˜}{b} :$

\begin{matrix} (i) - - \to A P (ii) - - \to O Q \end{matrix}

$\begin{array}{l} (i) \vec{A P} \\ (ii) \vec{O Q} \end{array}$

(b)(i) Given that

- - \to A R = h - - \to A P

$\vec{A R} = h \vec{A P}$ , state

- - \to A R

$\vec{A R}$ in terms of h,

a ˜ and b ˜ .

$\underset{˜}{a} and \underset{˜}{b} .$
(ii) Given that

- - \to R Q = k - - \to O Q,

$\vec{R Q} = k \vec{O Q},$ state in terms of k,

a ˜ and b ˜ .

$\underset{˜}{a} and \underset{˜}{b} .$

(c) Using

- - \to A Q = - - \to A R + - - \to R Q,

$\vec{A Q} = \vec{A R} + \vec{R Q},$ find the value of h and of k.

Solution:
(a)(i)

\begin{matrix} - - \to A P = - - \to A O + - - \to O P - - \to A P = - - - \to O A + - - \to O P - - \to A P = - 8 a ˜ + 4 b ˜ \end{matrix}

$\begin{array}{l} \vec{A P} = \vec{A O} + \vec{O P} \\ \vec{A P} = - \vec{O A} + \vec{O P} \\ \vec{A P} = - 8 \underset{˜}{a} + 4 \underset{˜}{b} \end{array}$

(a)(ii)

\begin{matrix} - - \to O Q = - - \to O A + - - \to A Q - - \to O Q = 8 a ˜ + \frac{1}{4} - - \to A B - - \to O Q = 8 a ˜ + \frac{1}{4} (- - \to A O + - - \to O B) - - \to O Q = 8 a ˜ + \frac{1}{4} (- 8 a ˜ + 4 - - \to O P) - - \to O Q = 8 a ˜ + \frac{1}{4} (- 8 a ˜ + 4 (4 b ˜)) - - \to O Q = 8 a ˜ - 2 a ˜ + 4 b ˜ - - \to O Q = 6 a ˜ + 4 b ˜ \end{matrix}

$\begin{array}{l} \vec{O Q} = \vec{O A} + \vec{A Q} \\ \vec{O Q} = 8 \underset{˜}{a} + \frac{1}{4} \vec{A B} \\ \vec{O Q} = 8 \underset{˜}{a} + \frac{1}{4} (\vec{A O} + \vec{O B}) \\ \vec{O Q} = 8 \underset{˜}{a} + \frac{1}{4} (- 8 \underset{˜}{a} + 4 \vec{O P}) \\ \vec{O Q} = 8 \underset{˜}{a} + \frac{1}{4} (- 8 \underset{˜}{a} + 4 (4 \underset{˜}{b})) \\ \vec{O Q} = 8 \underset{˜}{a} - 2 \underset{˜}{a} + 4 \underset{˜}{b} \\ \vec{O Q} = 6 \underset{˜}{a} + 4 \underset{˜}{b} \end{array}$

(b)(i)

\begin{matrix} - - \to A R = h - - \to A P - - \to A R = h (- 8 a ˜ + 4 b ˜) - - \to A R = - 8 h a ˜ + 4 h b ˜ \end{matrix}

$\begin{array}{l} \vec{A R} = h \vec{A P} \\ \vec{A R} = h (- 8 \underset{˜}{a} + 4 \underset{˜}{b}) \\ \vec{A R} = - 8 h \underset{˜}{a} + 4 h \underset{˜}{b} \end{array}$

(b)(ii)

\begin{matrix} - - \to R Q = k - - \to O Q - - \to R Q = k (6 a ˜ + 4 b ˜) - - \to R Q = 6 k a ˜ + 4 k b ˜ \end{matrix}

$\begin{array}{l} \vec{R Q} = k \vec{O Q} \\ \vec{R Q} = k (6 \underset{˜}{a} + 4 \underset{˜}{b}) \\ \vec{R Q} = 6 k \underset{˜}{a} + 4 k \underset{˜}{b} \end{array}$

(c)

\begin{matrix} - - \to A Q = - - \to A R + - - \to R Q - - \to A Q = - 8 h a ˜ + 4 h b ˜ + (6 k a ˜ + 4 k b ˜) - - \to A O + - - \to O Q = - 8 h a ˜ + 4 h b ˜ + 6 k a ˜ + 4 k b ˜ - 8 a ˜ + 6 a ˜ + 4 b ˜ = - 8 h a ˜ + 6 k a ˜ + 4 h b ˜ + 4 k b ˜ - 2 a ˜ + 4 b ˜ = - 8 h a ˜ + 6 k a ˜ + 4 h b ˜ + 4 k b ˜ - 2 = - 8 h + 6 k - 1 = - 4 h + 3 k \to (1) 4 = 4 h + 4 k 1 = h + k k = 1 - h \to (2) Substitute (2) into (1), - 1 = - 4 h + 3 (1 - h) - 1 = - 4 h + 3 - 3 h - 4 = - 7 h h = \frac{4}{7} From (2), k = 1 - \frac{4}{7} = \frac{3}{7} \end{matrix}

$\begin{array}{l} \vec{A Q} = \vec{A R} + \vec{R Q} \\ \vec{A Q} = - 8 h \underset{˜}{a} + 4 h \underset{˜}{b} + (6 k \underset{˜}{a} + 4 k \underset{˜}{b}) \\ \vec{A O} + \vec{O Q} = - 8 h \underset{˜}{a} + 4 h \underset{˜}{b} + 6 k \underset{˜}{a} + 4 k \underset{˜}{b} \\ - 8 \underset{˜}{a} + 6 \underset{˜}{a} + 4 \underset{˜}{b} = - 8 h \underset{˜}{a} + 6 k \underset{˜}{a} + 4 h \underset{˜}{b} + 4 k \underset{˜}{b} \\ - 2 \underset{˜}{a} + 4 \underset{˜}{b} = - 8 h \underset{˜}{a} + 6 k \underset{˜}{a} + 4 h \underset{˜}{b} + 4 k \underset{˜}{b} \\ - 2 = - 8 h + 6 k \\ - 1 = - 4 h + 3 k \to (1) \\ 4 = 4 h + 4 k \\ 1 = h + k \\ k = 1 - h \to (2) \\ Substitute (2) into (1), \\ - 1 = - 4 h + 3 (1 - h) \\ - 1 = - 4 h + 3 - 3 h \\ - 4 = - 7 h \\ h = \frac{4}{7} \\ From (2), \\ k = 1 - \frac{4}{7} = \frac{3}{7} \end{array}$

Question 2:
Given that

- - \to A B = (\begin{matrix} 1014 \end{matrix}), - - \to O B = (\begin{matrix} 46 \end{matrix})

$\vec{A B} = (\begin{matrix} 10 \\ 14 \end{matrix}), \vec{O B} = (\begin{matrix} 4 \\ 6 \end{matrix})$ and

- - \to C D = (\begin{matrix} m 7 \end{matrix})

$\vec{C D} = (\begin{matrix} m \\ 7 \end{matrix})$ , find
(a) the coordinates of A,
(b) the unit vector in the direction of

- - \to O A

$\vec{O A}$ .
(c) the value of m if CD is parallel to AB .

Solution:
(a)

\begin{matrix} - - \to A B = - - \to A O + - - \to O B (\begin{matrix} 1014 \end{matrix}) = (\begin{matrix} x y \end{matrix}) + (\begin{matrix} 46 \end{matrix}) (\begin{matrix} x y \end{matrix}) = (\begin{matrix} 1014 \end{matrix}) - (\begin{matrix} 46 \end{matrix}) - - \to A O = (\begin{matrix} 68 \end{matrix}) - - \to O A = (\begin{matrix} - 6 - 8 \end{matrix}) A = (- 6, - 8) \end{matrix}

$\begin{array}{l} \vec{A B} = \vec{A O} + \vec{O B} \\ (\begin{matrix} 10 \\ 14 \end{matrix}) = (\begin{matrix} x \\ y \end{matrix}) + (\begin{matrix} 4 \\ 6 \end{matrix}) \\ (\begin{matrix} x \\ y \end{matrix}) = (\begin{matrix} 10 \\ 14 \end{matrix}) - (\begin{matrix} 4 \\ 6 \end{matrix}) \\ \vec{A O} = (\begin{matrix} 6 \\ 8 \end{matrix}) \\ \vec{O A} = (\begin{matrix} - 6 \\ - 8 \end{matrix}) \\ A = (- 6, - 8) \end{array}$

(b)

\begin{matrix} ∣ ∣ ∣ - - \to O A ∣ ∣ ∣ = \sqrt{{(- 6)}^{2} + {(- 8)}^{2}} ∣ ∣ ∣ - - \to O A ∣ ∣ ∣ = \sqrt{100} = 10 the unit vector in the direction of - - \to O A = \frac{- - \to O A}{∣ ∣ ∣ - - \to O A ∣ ∣ ∣} = \frac{(\begin{matrix} - 6 - 8 \end{matrix})}{10} = \frac{1}{10} (\begin{matrix} - 6 - 8 \end{matrix}) = (\begin{matrix} - \frac{3}{5} - \frac{4}{5} \end{matrix}) \end{matrix}

$\begin{array}{l} | \vec{O A} | = \sqrt{{(- 6)}^{2} + {(- 8)}^{2}} \\ | \vec{O A} | = \sqrt{100} = 10 \\ the unit vector in the direction of \vec{O A} \\ = \frac{\vec{O A}}{| \vec{O A} |} \\ = \frac{(\begin{matrix} - 6 \\ - 8 \end{matrix})}{10} = \frac{1}{10} (\begin{matrix} - 6 \\ - 8 \end{matrix}) \\ = (\begin{matrix} - \frac{3}{5} \\ - \frac{4}{5} \end{matrix}) \end{array}$

(c)

$\begin{array}{l} Given \vec{C D} parallel \vec{A B} \\ ∴ \vec{C D} = k \vec{A B} \\ (\begin{matrix} m \\ 7 \end{matrix}) = k (\begin{matrix} 10 \\ 14 \end{matrix}) \\ (\begin{matrix} m \\ 7 \end{matrix}) = (\begin{matrix} 10 k \\ 14 k \end{matrix}) \\ 7 = 14 k \\ k = \frac{1}{2} \\ m = 10 k = 10 (\frac{1}{2}) = 5 \end{array}$

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