Question

A plane \( \pi_1 \) passing through the point \( 3\hat{i} - 7\hat{j} + 5\hat{k} \) is perpendicular to the vector \( \hat{i} + 2\hat{j} - 2\hat{k} \) and another plane \( \pi_2 \) passing through the point \( 2\hat{i} + 7\hat{j} - 8\hat{k} \) is perpendicular to the vector \( 3\hat{i} + 2\hat{j} + 6\hat{k} \). If \( p_1 \) and \( p_2 \) are the perpendicular distances from the origin to the planes \( \pi_1 \) and \( \pi_2 \) respectively, then \( p_1 - p_2 \) is:

• A. \( 1 \)
• B. \( 2 \)
• C. \( 3 \)
• D. \( 4 \)

Hint:

To find the perpendicular distance from the origin to a plane, use the formula \( \frac{|D|}{\sqrt{A^2 + B^2 + C^2}} \), where \( D \) is the constant term in the plane equation, and \( A, B, C \) are the coefficients of the variables.

Answer 1

The Correct Option is C

We are given two planes: - Plane \( \pi_1 \): passing through the point \( P_1(3\hat{i} - 7\hat{j} + 5\hat{k}) \) with normal vector \( \hat{i} + 2\hat{j} - 2\hat{k} \). - Plane \( \pi_2 \): passing through the point \( P_2(2\hat{i} + 7\hat{j} - 8\hat{k}) \) with normal vector \( 3\hat{i} + 2\hat{j} + 6\hat{k} \). Our goal is to determine the difference in perpendicular distances from the origin to these two planes. 

Step 1: Determine the equation of plane \( \pi_1 \). The general equation of a plane is: \[ Ax + By + Cz + D = 0 \] where \( A, B, C \) are the components of the normal vector, and \( D \) is a constant. For plane \( \pi_1 \), the normal vector is \( \hat{i} + 2\hat{j} - 2\hat{k} \), so its equation is: \[ x + 2y - 2z + D = 0. \] Substituting point \( P_1(3, -7, 5) \): \[ 3(1) + (-7)(2) + 5(-2) + D = 0 \] \[ 3 - 14 - 10 + D = 0 \] \[ D = 21 \] Thus, the equation of plane \( \pi_1 \) is: \[ x + 2y - 2z + 21 = 0. \] 

Step 2: Determine the equation of plane \( \pi_2 \). For plane \( \pi_2 \), the normal vector is \( 3\hat{i} + 2\hat{j} + 6\hat{k} \), so its equation is: \[ 3x + 2y + 6z + D = 0. \] Substituting point \( P_2(2, 7, -8) \): \[ 3(2) + 2(7) + 6(-8) + D = 0 \] \[ 6 + 14 - 48 + D = 0 \] \[ D = 28 \] Thus, the equation of plane \( \pi_2 \) is: \[ 3x + 2y + 6z + 28 = 0. \] 

Step 3: Compute the perpendicular distance from the origin to each plane. The formula for the perpendicular distance from the origin to the plane \( Ax + By + Cz + D = 0 \) is: \[ \text{Distance} = \frac{|D|}{\sqrt{A^2 + B^2 + C^2}}. \] For plane \( \pi_1 \) where \( A = 1, B = 2, C = -2, D = 21 \): \[ p_1 = \frac{|21|}{\sqrt{1^2 + 2^2 + (-2)^2}} = \frac{21}{\sqrt{1 + 4 + 4}} = \frac{21}{3} = 7. \] For plane \( \pi_2 \) where \( A = 3, B = 2, C = 6, D = 28 \): \[ p_2 = \frac{|28|}{\sqrt{3^2 + 2^2 + 6^2}} = \frac{28}{\sqrt{9 + 4 + 36}} = \frac{28}{7} = 4. \] 

Step 4: Compute the difference in distances. \[ p_1 - p_2 = 7 - 4 = 3. \] Thus, the required difference in perpendicular distances is: \[ \boxed{3}.