Question

The coordinates of the foot of the perpendicular drawn from a point $P(-1, 1, 2)$ to the plane $2x - 3y + z - 11 = 0$ are

• A. $(2, -2, 1)$
• B. $(2, -3, 0)$
• C. $(1, -2, 3)$
• D. $(4, 1, 6)$

Hint:

You can quickly check the answer by verifying if the point satisfies the plane equation and the vector $PF$ is parallel to the normal.

Answer 1

The Correct Option is C

Step 1: Concept
The coordinates $(x, y, z)$ of the foot of the perpendicular from $(x_1, y_1, z_1)$ to $ax + by + cz + d = 0$ satisfy $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} = -\frac{(ax_1 + by_1 + cz_1 + d)}{a^2 + b^2 + c^2}$.

Step 2: Meaning
Here $(x_1, y_1, z_1) = (-1, 1, 2)$ and $a=2, b=-3, c=1, d=-11$.

Step 3: Analysis
Ratio $= -\frac{[2(-1) - 3(1) + 2 - 11]}{2^2 + (-3)^2 + 1^2} = -\frac{-14}{14} = 1$. So, $\frac{x+1}{2} = 1 \implies x=1$; $\frac{y-1}{-3} = 1 \implies y=-2$; $\frac{z-2}{1} = 1 \implies z=3$.

Step 4: Conclusion
The foot of the perpendicular is $(1, -2, 3)$. Final Answer: (C)