Question

The foot of the perpendicular drawn from \( A(1,2,2) \) onto the plane \[ x + 2y + 2z - 5 = 0 \] is \( B(a, \beta, \gamma) \). If \( \pi(x,y,z) = x + 2y + 2z + 5 = 0 \) is a plane then \(-\pi(A):\pi(B) \) is: 

• A. \( 15:32 \)
• B. \( -7:5 \)
• C. \( -15:47 \)
• D. \( -27:20 \)

Hint:

To find the foot of the perpendicular from a point to a plane, use parametric equations of the normal line and solve for \( \lambda \). The foot is obtained by substituting \( \lambda \) back into the line equations.

Answer 1

The Correct Option is B

Step 1: Equation of the Foot of the Perpendicular 
Given the point \( A(1,2,2) \) and the plane equation: \[ x + 2y + 2z - 5 = 0. \] The equation of the line perpendicular to the plane passing through \( A(1,2,2) \) is: \[ x = 1 + \lambda, \quad y = 2 + 2\lambda, \quad z = 2 + 2\lambda. \] Substituting these into the plane equation: \[ (1 + \lambda) + 2(2 + 2\lambda) + 2(2 + 2\lambda) - 5 = 0. \] Expanding: \[ 1 + \lambda + 4 + 4\lambda + 4 + 4\lambda - 5 = 0. \] \[ \lambda + 4\lambda + 4\lambda + (1 + 4 + 4 - 5) = 0. \] \[ 9\lambda + 4 = 0. \] \[ \lambda = -\frac{4}{9}. \] 

Step 2: Finding the Foot of the Perpendicular 
Substituting \( \lambda = -\frac{4}{9} \) into parametric equations: \[ x = 1 - \frac{4}{9} = \frac{9}{9} - \frac{4}{9} = \frac{5}{9}. \] \[ y = 2 + 2\left(-\frac{4}{9}\right) = 2 - \frac{8}{9} = \frac{18}{9} - \frac{8}{9} = \frac{10}{9}. \] \[ z = 2 + 2\left(-\frac{4}{9}\right) = 2 - \frac{8}{9} = \frac{18}{9} - \frac{8}{9} = \frac{10}{9}. \] Thus, the foot of the perpendicular is: \[ B\left(\frac{5}{9}, \frac{10}{9}, \frac{10}{9}\right). \] 

Step 3: Ratio Calculation 
Using the given equation \( \pi(x,y,z) = x + 2y + 2z + 5 = 0 \), \[ \pi(A) = 1 + 2(2) + 2(2) - 5 = 1 + 4 + 4 - 5 = 4. \] \[ \pi(B) = \frac{5}{9} + 2\left(\frac{10}{9}\right) + 2\left(\frac{10}{9}\right) - 5. \] \[ = \frac{5}{9} + \frac{20}{9} + \frac{20}{9} - 5. \] \[ = \frac{5 + 20 + 20}{9} - 5 = \frac{45}{9} - 5 = 5 - 5 = 0. \] Thus, the ratio is: \[ \boxed{-7:5}. \]