Question

Let \( \pi \) be the plane that passes through the point \( (-2,1,-1) \) and is parallel to the plane \( 2x - y + 2z = 0 \). Then the foot of the perpendicular drawn from the point \( (1,2,1) \) to the plane \( \pi \) is:

• A. \( (-3,-1,1) \)
• B. \( (-1,1,-3) \)
• C. \( (-3,3,-1) \)
• D. \( (-1,3,-1) \)

Hint:

To find the foot of the perpendicular from a point to a plane, use the parametric formula: \[ (x,y,z) = (x_1 - \lambda A, y_1 - \lambda B, z_1 - \lambda C). \] Solve for \( \lambda \) using the plane equation.

Answer 1

The Correct Option is D

Step 1: Finding the equation of the required plane
The given plane equation is: \[ 2x - y + 2z = 0. \] Since the required plane \( \pi \) is parallel to this plane, its equation must be of the form: \[ 2x - y + 2z = d. \] Since the plane passes through \( (-2,1,-1) \), substituting these values: \[ 2(-2) - 1(1) + 2(-1) = d. \] \[ -4 - 1 - 2 = d \Rightarrow d = -7. \] Thus, the equation of the required plane \( \pi \) is: \[ 2x - y + 2z = -7. \] Step 2: Finding the foot of the perpendicular
The formula for the foot of the perpendicular from \( (x_1, y_1, z_1) \) to the plane \( Ax + By + Cz + D = 0 \) is: \[ x = x_1 - \lambda A, \quad y = y_1 - \lambda B, \quad z = z_1 - \lambda C. \] Substituting \( A = 2 \), \( B = -1 \), \( C = 2 \), and the given point \( (1,2,1) \): \[ x = 1 - \lambda(2), \quad y = 2 - \lambda(-1), \quad z = 1 - \lambda(2). \] Since the foot of the perpendicular lies on the plane \( 2x - y + 2z = -7 \), substituting these values: \[ 2(1 - 2\lambda) - (2 + \lambda) + 2(1 - 2\lambda) = -7. \] Expanding: \[ 2 - 4\lambda - 2 - \lambda + 2 - 4\lambda = -7. \] \[ -9\lambda + 2 = -7. \] \[ -9\lambda = -9 \quad \Rightarrow \quad \lambda = 1. \] Step 3: Finding the coordinates
\[ x = 1 - 2(1) = -1, \quad y = 2 + 1 = 3, \quad z = 1 - 2(1) = -1. \] Thus, the foot of the perpendicular is: \[ (-1,3,-1). \]