# 4.6.2 Vector Short Questions (Question 4 & 5)

Question 4:
Diagram below shows a parallelogram ABCD with BED as a straight line.

Solution:

(a)
$\begin{array}{l}\text{Note: for parallelogram,}\stackrel{\to }{AB}=\stackrel{\to }{DC}=7\underset{˜}{p},\text{}\stackrel{\to }{AD}=\stackrel{\to }{BC}=5\underset{˜}{q}.\\ \stackrel{\to }{BD}=\stackrel{\to }{BA}+\stackrel{\to }{AD}\\ \stackrel{\to }{BD}=-7\underset{˜}{p}+5\underset{˜}{q}\text{}\end{array}$

(b)

$\begin{array}{l}\stackrel{\to }{EC}=\stackrel{\to }{EB}+\stackrel{\to }{BC}\\ \stackrel{\to }{EC}=\frac{7}{4}\underset{˜}{p}-\frac{5}{4}\underset{˜}{q}+5\underset{˜}{q}\\ \stackrel{\to }{EC}=\frac{7}{4}\underset{˜}{p}+\frac{15}{4}\underset{˜}{q}\end{array}$

Question 5:

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

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