3D Geometry 3 Question 9
####9. A plane passing through the points $(0,-1,0)$ and $(0,0,1)$ and making an angle $\frac{\pi}{4}$ with the plane $y-z+5=0$, also passes through the point
(2019 Main, 9 April I)
(a) $(\sqrt{2}, 1,4)$
(b) $(-\sqrt{2}, 1,-4)$
(c) $(-\sqrt{2},-1,-4)$
(d) $(\sqrt{2},-1,4)$
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Answer:
(a)
Solution:
- Let the equation of plane is
$ a x+b y+c z=d $
Since plane (i) passes through the points $(0,-1,0)$ and $(0,0,1)$, then $-b=d$ and $c=d$
$\therefore$ Equation of plane becomes $a x-d y+d z=d$ $\because$ The plane (ii) makes an angle of $\frac{\pi}{4}$ with the plane
$ \begin{aligned} y-z+5 & =0 . \\ \cos \frac{\pi}{4} & =\left|\frac{-d-d}{\sqrt{a^{2}+d^{2}+d^{2}} \sqrt{1+1}}\right| \end{aligned} $
$[\because$ The angle between the two planes $a_{1} x+b_{1} y+c_{1} z+d=0$ and $a_{2} x+b_{2} y+c_{2} z+d=0$ is
$ \cos \theta=\left|\frac{a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}} \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}\right| $
$\Rightarrow \frac{1}{\sqrt{2}}=\frac{|-2 d|}{\sqrt{a^{2}+2 d^{2}} \sqrt{2}} \Rightarrow \sqrt{a^{2}+2 d^{2}}=|-d|$
$\Rightarrow a^{2}+2 d^{2}=4 d^{2}$
[squaring both sides]
$\Rightarrow a^{2}=2 d^{2} \Rightarrow a= \pm \sqrt{2} d$
So, the Eq. (ii) becomes
$ \pm \sqrt{2} x-y+z=1 $
Now, from options $(\sqrt{2}, 1,4)$ satisfy the plane
$ -\sqrt{2} x-y+z=1 $
