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}.