Question

A plane which bisects the angle between the two given planes \(x+2y+2z-2=0\) and \(2x-y+2z-4=0\), passes through the point

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

Hint:

For angle bisectors of two planes, use \(\frac{P_1}{|n_1|}=\pm\frac{P_2}{|n_2|}\).

Answer 1

The Correct Option is B

Concept:
The angle bisectors of two planes \[ P_1=0 \] and \[ P_2=0 \] are given by \[ \frac{P_1}{\sqrt{a_1^2+b_1^2+c_1^2}} = \pm \frac{P_2}{\sqrt{a_2^2+b_2^2+c_2^2}} \]

Step 1: Write the two planes. \[ P_1=x+2y+2z-2=0 \] \[ P_2=2x-y+2z-4=0 \] For the first plane, \[ \sqrt{1^2+2^2+2^2}=\sqrt{9}=3 \] For the second plane, \[ \sqrt{2^2+(-1)^2+2^2}=\sqrt{9}=3 \]

Step 2: Write the angle bisector condition. Since both denominators are equal, \[ P_1=\pm P_2 \] So the angle bisector planes satisfy \[ x+2y+2z-2=\pm(2x-y+2z-4) \]

Step 3: Test the correct point \((2,-4,1)\). For \[ (2,-4,1) \] calculate \(P_1\): \[ P_1=2+2(-4)+2(1)-2 \] \[ =2-8+2-2 \] \[ =-6 \] Now calculate \(P_2\): \[ P_2=2(2)-(-4)+2(1)-4 \] \[ =4+4+2-4 \] \[ =6 \] Thus, \[ P_1=-P_2 \] So this point lies on one of the angle bisector planes.

Step 4: Final answer. \[ \boxed{(2,-4,1)} \]