If the vectors $\vec{a}=2 \hat{i}+p \hat{j}+4 \hat{k}$ and $\vec{b}=6 \hat{i}-9 \hat{j}+q \hat{k}$ are…
If the vectors $\vec{a}=2 \hat{i}+p \hat{j}+4 \hat{k}$ and $\vec{b}=6 \hat{i}-9 \hat{j}+q \hat{k}$ are collinear, then $\mathrm{p}$ and $\mathrm{q}$ are
$\mathrm{p}=3, \mathrm{q}=-2$
$\mathrm{p}=3, \mathrm{q}=\mathrm{n} 12$
$\mathrm{p}=-3, \mathrm{q}=12$
$\mathrm{p}=-3, \mathrm{q}=-12$
Solution
Let $\vec{a}=x \vec{b}$
$\begin{aligned}
& \therefore 2 \hat{i}+p \hat{j}+4 \hat{k}=6 x \hat{i}-9 x \hat{j}+q x \hat{j} \\
& \therefore 2=6 x \Rightarrow x=\frac{1}{3} \\
& p=-9 x \Rightarrow(-9)\left(\frac{1}{3}\right)=-3 \text { and } 4=q x=q\left(\frac{1}{3}\right) \Rightarrow q=4(3)=12
\end{aligned}$