3D Geometry 3 Question 2
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) $(2,-4,1)$
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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 ( $-ve)$ 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 $3 x+y+4 z=6$
