Question

The foot of the perpendicular drawn from a point with position vector $\hat{i}+4\hat{k}$ on the line joining the points having position vectors as $-11\hat{j}+3\hat{k}$ and $2\hat{i}-3\hat{j}+\hat{k}$ has the position vector

• A. $4\hat{i}+5\hat{j}+5\hat{k}$
• B. $4\hat{i}+5\hat{j}-5\hat{k}$
• C. $5\hat{i}+4\hat{j}-5\hat{k}$
• D. None of these.

Answer 1

Answer: (D)

None of these.

Assume given point is $P(1,0,4)$
and equation of line passing through $(0,-11,3)$ and $(2,-3,1)$ is given by,
$\frac{x}{2}=\frac{y+11}{8}=\frac{z-3}{-2}=k$(say) $.......\left(1\right)$
So any point on this line is given as$Q(2k,8k-11,-2k+3)$
Now direction ratios of PQ are, $2k-1,8k-11,-2k-1$
Also $PQ\perp \left(1\right)$
$\Rightarrow 2(2k-1)+8(8k-11)-2(-2k-1)=0$
$\Rightarrow 72k-88=0$

$\Rightarrow k=\frac{11}{9}$
Hence, foot of perpendicular drawn on the given line is $\left(\frac{22}{9},\frac{-11}{9},\frac{5}{9}\right)$ or $\frac{22}{9}\hat{i}-\frac{11}{9}\hat{j}+\frac{5}{9}\hat{k}$