Question

If the plane $-4x-2y+2z+\alpha=0$ is at a distance of two units from the plane $2x+y-z+1=0$, then the product of all the possible values of $\alpha$ is

• A. -23
• B. 42
• C. -92
• D. 72

Hint:

Before applying the distance formula for parallel planes, make sure to rewrite the equations so that the coefficients of $x, y,$ and $z$ are exactly the same. This ensures you are using the correct values for $D_1$ and $D_2$.

Answer 1

The Correct Option is C

Step 1: Normalize the plane equations to check for parallelism.
Let $\pi_1: -4x-2y+2z+\alpha=0$. Dividing by -2, we get an equivalent equation: $2x+y-z-\frac{\alpha}{2}=0$. Let $\pi_2: 2x+y-z+1=0$. Since the coefficients of $x, y, z$ are identical, the planes are parallel.

Step 2: Use the formula for the distance between parallel planes.
The distance $d$ between two parallel planes $Ax+By+Cz+D_1=0$ and $Ax+By+Cz+D_2=0$ is given by: \[ d = \frac{|D_1-D_2|}{\sqrt{A^2+B^2+C^2}}. \] Here, $A=2, B=1, C=-1, D_1=-\alpha/2, D_2=1$. We are given $d=2$. \[ 2 = \frac{|-\frac{\alpha}{2} - 1|}{\sqrt{2^2+1^2+(-1)^2}} = \frac{|-\frac{\alpha+2}{2}|}{\sqrt{4+1+1}} = \frac{|\alpha+2|}{2\sqrt{6}}. \]

Step 3: Solve for the possible values of $\alpha$.
From the equation, we have $|\alpha+2| = 4\sqrt{6}$. This gives two possibilities: Case 1: $\alpha+2 = 4\sqrt{6} \implies \alpha_1 = 4\sqrt{6}-2$. Case 2: $\alpha+2 = -4\sqrt{6} \implies \alpha_2 = -4\sqrt{6}-2$.

Step 4: Calculate the product of the possible values of $\alpha$.
We need to find the product $\alpha_1 \cdot \alpha_2$. \[ \alpha_1 \alpha_2 = (4\sqrt{6}-2)(-4\sqrt{6}-2) = (-2+4\sqrt{6})(-2-4\sqrt{6}). \] This is in the form $(u+v)(u-v) = u^2-v^2$, where $u=-2$ and $v=4\sqrt{6}$. \[ \text{Product} = (-2)^2 - (4\sqrt{6})^2 = 4 - (16 \times 6) = 4 - 96 = -92. \] \[ \boxed{\alpha_1 \alpha_2 = -92}. \]