Vectors

Question 1:

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

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

(a) Express in terms of

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

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

(b)(i) Given that

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

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

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

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

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

(c) Using

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

Solution:
(a)(i)

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

(a)(ii)

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

(b)(i)

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

(b)(ii)

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

(c)

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

Question 3:

In diagram below, PQRS is a quadrilateral. PTS and TUR are straight lines.

It is given that

$\vec{P Q} = 20 \underset{˜}{x}, \vec{P T} = 8 \underset{˜}{y}, \vec{S R} = 25 \underset{˜}{x} - 24 \underset{˜}{y}, \vec{P T} = \frac{1}{4} \vec{P S} and \vec{T U} = \frac{3}{5} \vec{T R}$

(a) Express in terms of

$\underset{˜}{x}$ and/or

$\underset{˜}{y}$ :

(i) $\vec{Q S}$
(ii) $\vec{T R}$
(b) Show that the points Q, U and S are collinear.

(c) If

$| \underset{˜}{x} |$ = 2 and

$| \underset{˜}{y} |$ = 3, find

$| \vec{Q S} |$

Solution:
(a)(i)

$\begin{array}{l} \vec{Q S} = \vec{Q P} + \vec{P S} \\ \vec{Q S} = - 20 \underset{˜}{x} + 32 \underset{˜}{y} \leftarrow \begin{array}{l} Given \vec{P T} = \frac{1}{4} \vec{P S} \\ \vec{P S} = 4 \vec{P T} = 4 (8 \underset{˜}{y}) = 32 \underset{˜}{y} \end{array} \end{array}$

(a)(ii)

$\begin{array}{l} \vec{T R} = \vec{T S} + \vec{S R} \\ \vec{T R} = \frac{3}{4} \vec{P S} + 25 \underset{˜}{x} - 24 \underset{˜}{y} \\ \vec{T R} = \frac{3}{4} (32 \underset{˜}{y}) + 25 \underset{˜}{x} - 24 \underset{˜}{y} \\ \vec{T R} = 24 \underset{˜}{y} + 25 \underset{˜}{x} - 24 \underset{˜}{y} \\ \vec{T R} = 25 \underset{˜}{x} \end{array}$

(b)

$\begin{array}{l} \vec{Q U} = \vec{Q P} + \vec{P T} + \vec{T U} \\ \vec{Q U} = - 20 \underset{˜}{x} + 8 \underset{˜}{y} + \frac{3}{5} (25 \underset{˜}{x}) \leftarrow \begin{array}{l} Given \\ \vec{T U} = \frac{3}{5} \vec{T R} \end{array} \\ \vec{Q U} = - 20 \underset{˜}{x} + 8 \underset{˜}{y} + 15 \underset{˜}{x} \\ \vec{Q U} = - 5 \underset{˜}{x} + 8 \underset{˜}{y} \\ From (a)(i) \vec{Q S} = - 20 \underset{˜}{x} + 32 \underset{˜}{y} \\ \frac{\vec{Q S}}{\vec{Q U}} = \frac{- 20 \underset{˜}{x} + 32 \underset{˜}{y}}{- 5 \underset{˜}{x} + 8 \underset{˜}{y}} \\ \frac{\vec{Q S}}{\vec{Q U}} = \frac{4 (- 5 \underset{˜}{x} + 8 \underset{˜}{y})}{(- 5 \underset{˜}{x} + 8 \underset{˜}{y})} \\ \frac{\vec{Q S}}{\vec{Q U}} = 4 \\ \vec{Q S} = 4 \vec{Q U} \\ ∴ Q, U and S are collinear . \end{array}$

(c)

$\begin{array}{l} \vec{P S} = 32 \underset{˜}{y} \\ | \vec{P S} | = 32 | \underset{˜}{y} | \\ | \vec{P S} | = 32 \times 3 = 96 \\ \vec{P Q} = 20 \underset{˜}{x} \\ | \vec{P Q} | = 20 | \underset{˜}{x} | \\ | \vec{P Q} | = 20 \times 2 = 40 \\ ∴ | \vec{Q S} | = \sqrt{96^{2} + 40^{2}} \\ | \vec{Q S} | = 104 \end{array}$

Long Question 1

Long Question 3