Question

The co-ordinate of the foot of the perpendicular from $(1,8,4)$ to the line joining the points $(0,1,3),(2,-3,-1)$ is

• A. $\left(\frac{5}{3},\frac{-2}{3},\frac{19}{3}\right)$
• B. $\left(\frac{1}{3},\frac{2}{3},\frac{19}{3}\right)$
• C. $(-\frac{5}{3},\frac{2}{3}\frac{-19}{3})$
• D. $(-\frac{1}{3},\frac{2}{3},1)$

Answer 1

Answer: (C)

$(-\frac{5}{3},\frac{2}{3}\frac{-19}{3})$

Let $A(1,8,4),B(0,1,3)$ and $C(2,-3,-1)$ be any $3$ points on a plane.
Any point lying on the line $BC$ can be written in the form of $D(2k,1-4k.3-4k)$.
The dr's of $BC$ are $(2,-4,-4)$
The dr's of $AD$ are $(1-2k,7+4k,1+4k)$
Since, $AD$ has to be perpendicular to $BC$,
$AD.BC=0$

$\Rightarrow 2-4k-28-16k-4-16k=0$

$\Rightarrow k=\frac{-5}{6}$
Putting back the value of $k$, the point is

$(\frac{-5}{3},\frac{2}{3},\frac{-19}{3})$