Question

If three points A, B and C have position vectors $\hat{i}+x\hat{j}+3\hat{k},3\hat{i}+4\hat{j}+7\hat{k}\mathrm{and}y\hat{i}-2\hat{j}-5\hat{k}$ respectively are collinear, then (x, y) =
(a) (2, −3)
(b) (−2, 3)
(c) (−2, −3)
(d) (2, 3)

Answer 1

(a) (2, −3)
Given position vectors of $A,B$ and $C$ are $\hat{i}+x\hat{j}+3\hat{k},3\hat{i}+4\hat{j}+7\hat{k}$ and $y\hat{i}-2\hat{j}-5\hat{k}.$ Then,
$$\begin{gathered} \overrightarrow{AB}=3\hat{i}+4\hat{j}+7\hat{k}-\hat{i}-x\hat{j}-3\hat{k}=2\hat{i}+\left(4-x\right)\hat{j}+4\hat{k} \\ \overrightarrow{BC}=y\hat{i}-2\hat{j}-5\hat{k}-3\hat{i}-4\hat{j}-7\hat{k}=\left(y-3\right)\hat{i}-6\hat{j}-12\hat{k} \end{gathered}$$
Since, the given vectors are collinear.
$$\begin{gathered} \therefore \overrightarrow{AB}=\lambda \overrightarrow{BC} \\ \Rightarrow 2\hat{i}+\left(4-x\right)\hat{j}+4\hat{k}=\lambda \left(y-3\right)\hat{i}-6\lambda \hat{j}-12\lambda \hat{k} \\ \Rightarrow 2=\lambda \left(y-3\right),\left(4-x\right)=-6\lambda ,4=-12\lambda \\ \Rightarrow 2=\lambda \left(y-3\right),\left(4-x\right)=-6\lambda ,\lambda =-\frac{1}{3} \\ \Rightarrow 2=-\frac{1}{3}\left(y-3\right),\left(4-x\right)=-6\times \left(-\frac{1}{3}\right) \\ \Rightarrow -6=y-3,4-x=2 \\ \Rightarrow y=-3,x=2 \end{gathered}$$