# 4.7.1 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

(a) Express in terms of
$\begin{array}{l}\left(i\right)\text{}\stackrel{\to }{AP}\\ \text{(ii)}\stackrel{\to }{OQ}\end{array}$

(b)(i) Given that $\stackrel{\to }{AR}=h\stackrel{\to }{AP}$ , state   $\stackrel{\to }{AR}$  in terms of h
(ii) Given that   $\stackrel{\to }{RQ}=k\stackrel{\to }{OQ},$ state  in terms of k,

(c) Using   find the value of h and of k.

Solution
:

(a)(i)
$\begin{array}{l}\stackrel{\to }{AP}=\stackrel{\to }{AO}+\stackrel{\to }{OP}\\ \stackrel{\to }{AP}=-\stackrel{\to }{OA}+\stackrel{\to }{OP}\\ \stackrel{\to }{AP}=-8\underset{˜}{a}+4\underset{˜}{b}\end{array}$

(a)(ii)
$\begin{array}{l}\stackrel{\to }{OQ}=\stackrel{\to }{OA}+\stackrel{\to }{AQ}\\ \stackrel{\to }{OQ}=8\underset{˜}{a}+\frac{1}{4}\stackrel{\to }{AB}\\ \stackrel{\to }{OQ}=8\underset{˜}{a}+\frac{1}{4}\left(\stackrel{\to }{AO}+\stackrel{\to }{OB}\right)\\ \stackrel{\to }{OQ}=8\underset{˜}{a}+\frac{1}{4}\left(-8\underset{˜}{a}+4\stackrel{\to }{OP}\right)\\ \stackrel{\to }{OQ}=8\underset{˜}{a}+\frac{1}{4}\left(-8\underset{˜}{a}+4\left(4\underset{˜}{b}\right)\right)\\ \stackrel{\to }{OQ}=8\underset{˜}{a}-2\underset{˜}{a}+4\underset{˜}{b}\\ \stackrel{\to }{OQ}=6\underset{˜}{a}+4\underset{˜}{b}\end{array}$

(b)(i)
$\begin{array}{l}\stackrel{\to }{AR}=h\stackrel{\to }{AP}\\ \stackrel{\to }{AR}=h\left(-8\underset{˜}{a}+4\underset{˜}{b}\right)\\ \stackrel{\to }{AR}=-8h\underset{˜}{a}+4h\underset{˜}{b}\end{array}$

(b)(ii)
$\begin{array}{l}\stackrel{\to }{RQ}=k\stackrel{\to }{OQ}\\ \stackrel{\to }{RQ}=k\left(6\underset{˜}{a}+4\underset{˜}{b}\right)\\ \stackrel{\to }{RQ}=6k\underset{˜}{a}+4k\underset{˜}{b}\end{array}$

(c)
$\begin{array}{l}\stackrel{\to }{AQ}=\stackrel{\to }{AR}+\stackrel{\to }{RQ}\\ \stackrel{\to }{AQ}=-8h\underset{˜}{a}+4h\underset{˜}{b}+\left(6k\underset{˜}{a}+4k\underset{˜}{b}\right)\\ \stackrel{\to }{AO}+\stackrel{\to }{OQ}=-8h\underset{˜}{a}+4h\underset{˜}{b}+6k\underset{˜}{a}+4k\underset{˜}{b}\\ -8\underset{˜}{a}+6\underset{˜}{a}+4\underset{˜}{b}=-8h\underset{˜}{a}+6k\underset{˜}{a}+4h\underset{˜}{b}+4k\underset{˜}{b}\\ -2\underset{˜}{a}+4\underset{˜}{b}=-8h\underset{˜}{a}+6k\underset{˜}{a}+4h\underset{˜}{b}+4k\underset{˜}{b}\\ \\ -2=-8h+6k\\ -1=-4h+3k\to \left(1\right)\\ \\ 4=4h+4k\\ 1=h+k\\ k=1-h\to \left(2\right)\\ \\ \text{Substitute (2) into (1),}\\ -1=-4h+3\left(1-h\right)\\ -1=-4h+3-3h\\ -4=-7h\\ h=\frac{4}{7}\\ \\ \text{From (2),}\\ k=1-\frac{4}{7}=\frac{3}{7}\end{array}$

Question 2:
Given that   $\stackrel{\to }{AB}=\left(\begin{array}{c}10\\ 14\end{array}\right),\text{}\stackrel{\to }{OB}=\left(\begin{array}{c}4\\ 6\end{array}\right)$ and $\stackrel{\to }{CD}=\left(\begin{array}{c}m\\ 7\end{array}\right)$ , find
(a) the coordinates of A,
(b) the unit vector in the direction of $\stackrel{\to }{OA}$ .
(c) the value of m if CD is parallel to AB .

Solution:

(a)
$\begin{array}{l}\stackrel{\to }{AB}=\stackrel{\to }{AO}+\stackrel{\to }{OB}\\ \left(\begin{array}{c}10\\ 14\end{array}\right)=\left(\begin{array}{c}x\\ y\end{array}\right)+\left(\begin{array}{c}4\\ 6\end{array}\right)\\ \left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}10\\ 14\end{array}\right)-\left(\begin{array}{c}4\\ 6\end{array}\right)\\ \stackrel{\to }{AO}=\left(\begin{array}{c}6\\ 8\end{array}\right)\\ \stackrel{\to }{OA}=\left(\begin{array}{c}-6\\ -8\end{array}\right)\\ A=\left(-6,-8\right)\end{array}$

(b)

(c)