Question

If the distance between the planes \( 2x + y + z + 1 = 0 \) and \( 2x + y + z + \alpha = 0 \) is 3 units, then the product of all possible values of \( \alpha \) is:

• A. \( -43 \)
• B. \( 43 \)
• C. \( 53 \)
• D. \( -53 \)

Hint:

For distance between parallel planes, use the absolute difference of constants divided by the magnitude of the normal vector.

Answer 1

The Correct Option is D

Step 1: Using the distance formula between parallel planes The formula for the distance between two parallel planes \( Ax + By + Cz + D_1 = 0 \) and \( Ax + By + Cz + D_2 = 0 \) is: \[ \text{Distance} = \frac{|D_1 - D_2|}{\sqrt{A^2 + B^2 + C^2}} \] Substituting values: \[ 3 = \frac{|\alpha - 1|}{\sqrt{2^2 + 1^2 + 1^2}} \] \[ 3 = \frac{|\alpha - 1|}{\sqrt{6}} \] Solving for \( \alpha \), we get two values whose product is: \[ \alpha_1 \times \alpha_2 = -53 \]