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 $\hat{ i }+2 \hat{ j }-3 k$ 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 $P A=18$
$\therefore (\sqrt{2} \lambda+1-1)^{2}+(-5 \lambda+2-2)^{2}$
$+(3 \lambda-3+3)^{2}=18^{2}$
$2 \lambda^{2}+25 \lambda^{2}+9 \lambda^{2}=18^{2}$
$\lambda^{2}=\frac{18^{2}}{36}=9$
$\lambda=3,-3$
$\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 }$