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  4. The foot of the perpendicular from the point (1,6,3) to the line (x/1)=(y-1/2)=(z-2/3) is

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Q. The foot of the perpendicular from the point
$(1,6,3)$ to the line $\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}$ is

KEAMKEAM 2015Three Dimensional Geometry

Solution:

Any point on the line can be written as
$x=\lambda, y=2 \lambda+1, z=3 \lambda+2$
image
Hence, coordinates of
$Q$ are $(\lambda, 2 \lambda+1,3 \lambda+2) .$
DR's of
$P Q$ are $\lambda-1,2 \lambda+1-6,3 \lambda+2-3$
i.e. $\lambda-1,2 \lambda-5,3 \lambda-1$
Since, $P Q$ is perpendicular to the given lines.
So,1 $(\lambda-1)+2(2 \lambda-5)+3(3 \lambda-1)=0$
$\Rightarrow \lambda-1+4 \lambda-10+9 \lambda-3=0$
$\Rightarrow 14 \lambda=14 \Rightarrow \lambda=1$
$\therefore $ Coordinates of $Q=(1,3,5)$