Question

The coordinates of a point on the line $\frac{x-1}{2}=\frac{y+1}{-3}=z$ at a distance of $4\sqrt{14}$ from the point $(1,-1,0)$ nearer to the origin is :

• A. $(9,-13,4)$
• B. $(-7,11,-4)$
• C. $(-7,13,4)$
• D. $(-1,6,-4)$

Answer 1

The correct option is B $(-7,11,-4)$

Let $A(1,-1,0)$ be a point on given line $\frac{x-1}{2}=\frac{y+1}{-3}=z$
Assuming $\frac{x-1}{2}=\frac{y+1}{-3}=z=t$,
The coordinates of any point $P$ on the given line can be $(2t+1,-3t-1,t)$
Given, $|\overrightarrow{AP}|=4\sqrt{14}$
$\Rightarrow (4\sqrt{14}{)}^2=(2t+1-1{)}^2+(-3t-1+1{)}^2+(t{)}^2$
$\Rightarrow (4\sqrt{14}{)}^2=(2t{)}^2+(-3t{)}^2+(t{)}^2$
$\Rightarrow t=\pm 4$
With the corresponding values of $t$ we have two possible values of $P$ which are $(9,-13,4)$ and $(-7,11,-4)$ out of which, the nearest to origin is $(-7,11,-4)$