Question

If plane $P_1: x+y+z=k$ is one of the angle bisector plane of the planes $P_2: x+2y-6z=1$ and $P_3: 3x+4y-4z=9$, then k is equal to

• A. 4
• B. -4
• C. 5
• D. -5

Answer 1

Answer: (A)

4

Answer 2

The equations of the angle bisector planes of two planes $A_1x + B_1y + C_1z + D_1 = 0$ and $A_2x + B_2y + C_2z + D_2 = 0$ are given by: $\frac{A_1x + B_1y + C_1z + D_1}{\sqrt{A_1^2 + B_1^2 + C_1^2}} = \pm \frac{A_2x + B_2y + C_2z + D_2}{\sqrt{A_2^2 + B_2^2 + C_2^2}}$

The given planes are $P_2: x+2y-6z=1$ and $P_3: 3x+4y-4z=9$. Rewrite them in the standard form $Ax+By+Cz+D=0$: $P_2: x+2y-6z-1=0$ $P_3: 3x+4y-4z-9=0$

Calculate the magnitudes of the normal vectors: For $P_2$, $|N_2| = \sqrt{1^2 + 2^2 + (-6)^2} = \sqrt{1 + 4 + 36} = \sqrt{41}$. For $P_3$, $|N_3| = \sqrt{3^2 + 4^2 + (-4)^2} = \sqrt{9 + 16 + 16} = \sqrt{41}$.

Since $|N_2| = |N_3|$, the equations for the angle bisector planes are: $(x+2y-6z-1) = \pm (3x+4y-4z-9)$

Case 1: Using the '+' sign $x+2y-6z-1 = 3x+4y-4z-9$ $0 = (3x-x) + (4y-2y) + (-4z - (-6z)) + (-9 - (-1))$ $0 = 2x + 2y + 2z - 8$ Dividing by 2: $x+y+z-4 = 0$ $x+y+z = 4$

This equation matches the form $P_1: x+y+z=k$, so $k=4$.

Case 2: Using the '-' sign $x+2y-6z-1 = -(3x+4y-4z-9)$ $x+2y-6z-1 = -3x-4y+4z+9$ $(x+3x) + (2y+4y) + (-6z-4z) + (-1-9) = 0$ $4x+6y-10z-10 = 0$ Dividing by 2: $2x+3y-5z-5 = 0$ $2x+3y-5z = 5$

This equation does not match the form $x+y+z=k$.

Therefore, the plane $P_1: x+y+z=k$ is the angle bisector plane $x+y+z=4$. Thus, $k=4$.