Question

Let $L$ be the line parallel to the vector $\sqrt{2}\hat{i}-5\hat{j}+3\hat{k}$ and passing through the point
A given by $\hat{i}+2\hat{j}-3\hat{k}$. If the distance between $A$ and a point $P$ on the line $L$ is $18$ units, then the position vector of such a point $P$ is

• A. $(1-3\sqrt{2})\hat{i}+17\hat{j}-12\hat{k}$
• B. $(1+3\sqrt{2})\hat{i}+17\hat{j}+12\hat{k}$
• C. $(1+3\sqrt{2})\hat{i}-17\hat{j}-12\hat{k}$
• D. $(1-3\sqrt{2})\hat{i}-17\hat{j}+12\hat{k}$

Answer 1

Answer: (A)

(a) A line $L$ passing through $\hat{i}+2\hat{j}-3k$ and
parallel to vector $\sqrt{2}\hat{i}-5\hat{j}+3\hat{k}$ is
$r=\hat{i}+2\hat{j}-3\hat{k}+\lambda (\sqrt{2}\hat{i}-5\hat{j}+3\hat{k})$
Let any point $P$ lie on line is
$(\sqrt{2}\lambda +1,-5\lambda +2,3\lambda -3)$
point $A=\hat{i}+2\hat{j}-3\hat{k}$
Given $PA=18$
$\therefore (\sqrt{2}\lambda +1-1{)}^2+(-5\lambda +2-2{)}^2$
$+(3\lambda -3+3{)}^2=18^2$
$2{\lambda }^2+25{\lambda }^2+9{\lambda }^2=18^2$
${\lambda }^2=\frac{18^2}{36}=9$
$\lambda =3,-3$
$\therefore \text{ Point }P=(3\sqrt{2}+1)\hat{i}-13\hat{j}+6\hat{k}$
or $(1-3\sqrt{2})\hat{i}+17\hat{j}-12\hat{k}$