Question

If the points whose position vectors are $2\overline{i}+\overline{j}+\overline{k},6\overline{i}-\overline{j}+2\overline{k}$ and $14\overline{i}-5\overline{j}+p\overline{k}$ are collinear then the value of $p$ is

• A. $2$
• B. $4$
• C. $6$
• D. $8$

Answer 1

Answer: (B)

$4$

$2\hat{i}+\hat{j}+\hat{k}=\lambda (6\hat{i}-\hat{j}+2\hat{k})+\mu (14\hat{i}-5\hat{j}+p\hat{k})$ (Property of Collinear)
$2=b\lambda +14\mu$
$1=-\lambda -5\mu$
$0=8\lambda +24\mu$
$\frac{-\lambda }{3}=\mu$
$\lambda =\frac{3}{2}\mu =\frac{-1}{2}$
$1=2\lambda +p\mu$
$-2=\frac{-p}{2}$
$p=4$