Short Question 4 & 5

Question 4:

$\begin{array}{l}\text{Given that }\overrightarrow{AB}=7\underset{˜}{p},\overrightarrow{AD}=5\underset{˜}{q}\text{ and }DE=3EB,\text{ express, in terms of }\underset{˜}{p}\text{ and }\underset{˜}{q}.\\ \text{(a) }\overrightarrow{BD}\\ \text{(b) }\overrightarrow{EC}\end{array}$

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


(b)
$\begin{array}{l}\overrightarrow{DE}\text{=3}\overrightarrow{EB}\\ \frac{\overrightarrow{EB}}{\overrightarrow{DE}}=\frac{1}{3}\to EB:DE=1:3\\ \therefore \overrightarrow{EB}=\frac{1}{4}\overrightarrow{DB}\\ =\frac{1}{4}\left(-\overrightarrow{BD}\right)\\ =\frac{1}{4}\left[-\left(-7\underset{˜}{p}+5\underset{˜}{q}\right)\right]\leftarrow \text{From (a)}\\ \text{=}\frac{7}{4}\underset{˜}{p}+\frac{5}{4}\underset{˜}{q}\\ \\ \overrightarrow{EC}=\overrightarrow{EB}+\overrightarrow{BC}\\ \overrightarrow{EC}=\frac{7}{4}\underset{˜}{p}+\frac{5}{4}\underset{˜}{q}+5\underset{˜}{q}\\ \overrightarrow{EC}=\frac{7}{4}\underset{˜}{p}+\frac{25}{4}\underset{˜}{q}\end{array}$

Question 5:


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

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