Question

The orthogonal projection $A^{\prime }$ of the point $A$ with position vector $(1,2,3)$ on the plane $3x-y+4z=0$ is

• A. $(-1,3,-1)$
• B. $(-\frac{1}{2},\frac{5}{2},1)$
• C. $(\frac{1}{2},-\frac{5}{2},-1)$
• D. $(6,-7,-5)$

Answer 1

Answer: (B)

$(-\frac{1}{2},\frac{5}{2},1)$

Equation of line passes through $(1,2,3)$ and perpendicular to the given plane is given by,
$\frac{x-1}{3}=\frac{y-2}{-1}=\frac{z-3}{4}=k$ (say)
Let any point on this line is $P(3k+1,-k+2,4k+3)$
For orthogonal projection point $P$ lie on the given plane.
$\Rightarrow 3(3k+1)-(2-k)+4(4k+3)=0\Rightarrow k=-\frac{1}{2}$
Hence, required orthogonal point is $(\frac{-1}{2},\frac{5}{2},1)$