Question

The coordinates of the foot of perpendicular drawn from origin to the plane $2x-y+5z-3=0$ are ________.

• A. $(\frac{2}{\sqrt{30}},\frac{-1}{\sqrt{30}},\frac{5}{\sqrt{30}})$
• B. $(2,-1,5)$
• C. $(\frac{2}{3},\frac{-1}{3},\frac{5}{3})$
• D. $(\frac{1}{5},\frac{-1}{10},\frac{1}{2})$

Answer 1

Answer: (D)

$(\frac{1}{5},\frac{-1}{10},\frac{1}{2})$

The equation of the given plane is $2x-y+5z=3.$

So, the normal vector to this plane is $(2,-1,5).$

Therefore, the foot of the perpendicular to the given plane drawn from the origin will be of the form $(2t,-t,5t)$ for some $t\in R.$

Now, the foot of the perpendicular lies on the given plane. So it must satisfy the equation of the given plane.

Therefore, we solve for $t$ in

$2\left(2t\right)-(-t)+5\left(5t\right)=3.$

i.e, $4t+t+25t=3$

i.e, $t=\frac{1}{10}.$

So, the coordinates of the foot of the perpendicular to the given plane drawn from the origin is $(\frac{1}{5},\frac{-1}{10},\frac{1}{2}).$