Short Question 6 & 7

Question 6:
The points P, Q and R are collinear. It is given that   $\overrightarrow{PQ}=4\underset{˜}{a}-2\underset{˜}{b}$  and   $\overrightarrow{QR}=3\underset{˜}{a}+\left(1+k\right)\underset{˜}{b}$ , where k is a constant. Find
(a)    the value of k,
(b)    the ratio of PQ : QR.

Solution:
(a)
$\begin{array}{l}\text{Note: If }P,Q\text{ and }R\text{ are collinear,}\\ \overrightarrow{PQ}=m\overrightarrow{QR}\\ 4\underset{˜}{a}-2\underset{˜}{b}=m\left[3\underset{˜}{a}+\left(1+k\right)\underset{˜}{b}\right]\\ 4\underset{˜}{a}-2\underset{˜}{b}=3m\underset{˜}{a}+m\left(1+k\right)\underset{˜}{b}\\ \\ \text{Comparing vector:}\\ \underset{˜}{a}:4=3m\\ m=\frac{4}{3}\\ \underset{˜}{b}:-2=m\left(1+k\right)\\ -2=\frac{4}{3}\left(1+k\right)\\ 1+k=-\frac{6}{4}\\ k=-\frac{3}{2}-1\\ k=-\frac{5}{2}\end{array}$

(b)
$\begin{array}{l}\overrightarrow{PQ}=m\overrightarrow{QR}\\ \overrightarrow{PQ}=\frac{4}{3}\overrightarrow{QR}\\ \frac{\overrightarrow{PQ}}{\overrightarrow{QR}}=\frac{4}{3}\\ \therefore PQ:QR=4:3\end{array}$

Question 7:
Given that $\underset{˜}{x}=3\underset{˜}{i}+m\underset{˜}{j}$ and   $\underset{˜}{y}=4\underset{˜}{i}-3\underset{˜}{j}$ , find the values of m if the vector   $\underset{˜}{x}$  is parallel to the vector $\underset{˜}{y}$ .

Solution:
$\begin{array}{l}\text{If vector }\underset{˜}{x}\text{ is parallel to vector }\underset{˜}{y}\\ \underset{˜}{x}=h\underset{˜}{y}\\ \left(3\underset{˜}{i}+m\underset{˜}{j}\right)=h\left(4\underset{˜}{i}-3\underset{˜}{j}\right)\\ 3\underset{˜}{i}+m\underset{˜}{j}=4h\underset{˜}{i}-3h\underset{˜}{j}\\ \\ \text{Comparing vector:}\\ \underset{˜}{i}:3=4h\\ h=\frac{3}{4}\\ \underset{˜}{j}:m=-3h\\ m=-3\left(\frac{3}{4}\right)=-\frac{9}{4}\end{array}$