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

# A plane which bisects the angle between the two given planes and $x+2y+2z-2=0,$ passes through the point

1. A

$\left(2,4,1\right)$

2. B

$\left(1,4,-1\right)$

3. C

$\left(1,-4,1\right)$

4. D

$\left(2,-4,1\right)$

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### Solution:

Given planes are $2x-y+2z-4=0$ and

$x+2y+2z-2=0$

Equation of angle bisectors are

$\begin{array}{r}\frac{2x-y+2z-4}{\sqrt{4+1+4}}=±\left(\frac{x+2y+2z-2}{\sqrt{1+4+4}}\right)\\ ⇒\frac{2x-y+2z-4}{3}=±\left(\frac{x+2y+2z-2}{3}\right)\end{array}$

or $2x-y+2z-4=-\left(x+2y+2z-2\right)$

$⇒3x+y+4z-6=0$

Since, Point $\left(2,-4,1\right)$ satisfies equation (ii).

So, required point is $\left(2,-4,1\right)$.

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