3D Geometry 3 Question 3
####3. A plane which bisects the angle between the two given planes $2 x-y+2 z-4=0$ and $x+2 y+2 z-2=0$, passes through the point
(2019 Main, 12 April II)
(a) $(1,-4,1)$
(b) $(1,4,-1)$
(c) $(2,4,1)$
(d) $(2,-4,1)$
Show Answer
Answer:
(d)
Solution:
$ \begin{aligned} & \text { Key Idea Equation of planes bisecting the angles between the } \ & \text { planes } \ & a_{1} x+b_{1} y+c_{1} z+d_{1}=0 \text { and } \ & a_{2} x+b_{2} y+c_{2} z+d_{2}=0 \text {, are } \ & \frac{a_{1} x+b_{1} y+c_{1} z+d_{1}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}}}= \pm \frac{a_{2} x+b_{2} y+c_{2} z+d_{2}}{\sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}} \end{aligned} $
Equation of given planes are
$ \text { and } \quad \begin{aligned} 2 x-y+2 z-4 & =0 \ x+2 y+2 z-2 & =0 \end{aligned} $
Now, equation of planes bisecting the angles between the planes (i) and (ii) are
$ \begin{aligned} & \frac{2 x-y+2 z-4}{\sqrt{4+1+4}}= \pm \frac{x+2 y+2 z-2}{\sqrt{1+4+4}} \ & \Rightarrow \quad 2 x-y+2 z-4= \pm(x+2 y+2 z-2) \end{aligned} $
On taking (+ve) sign, we get a plane
$ x-3 y=2 $
On taking ( $-\mathrm{ve})$ sign, we get a plane
$ 3 x+y+4 z=6 $
Now from the given options, the point $(2,-4,1)$ satisfy the plane of angle bisector $3 x+y+4 z=6$
