For a,b Z and |a-b| 10, let the angle between the plane P:ax+y-z=b and the line l:x-1=a-y=…

Question

For $a,b\in Z$ and $|a-b|\le 10$, let the angle between the plane $P:ax+y-z=b$ and the line $l:x-1=a-y=z+1$ be ${\cos }^{-1}(\frac{1}{3})$. If the distance of the point $(6,-6,4)$ from the plane P is $3\sqrt{6}$, then ${a}^{4}+{b}^{2}$ is equal to :

Accepted Answer

We have, $	heta ={\cos }^{-1}\frac{1}{3}$

$\begin{gathered} \Rightarrow \cos 	heta =\frac{1}{3} \ 	herefore \sin 	heta =\sqrt{1-{(\frac{1}{3})}^{2}}=\sqrt{\frac{8}{9}}=\frac{2\sqrt{2}}{3} \end{gathered}$

The given plane line and are

$ax+y-z=b$

$x-1=a-y=z+1$

$\begin{gathered} 	herefore \sin 	heta =\frac{a\cdot 1+(1)(-1)+(-1)(1)}{\sqrt{{a}^{2}+{1}^{2}+{1}^{2}}\sqrt{{1}^{2}+{1}^{2}+{1}^{2}}} \ \Rightarrow \frac{a-1-1}{\sqrt{{a}^{2}+2}\sqrt{3}}=\frac{2\sqrt{2}}{3} \ \Rightarrow 3(a-2)=2\sqrt{6}\sqrt{{a}^{2}+2} \ \Rightarrow 9({a}^{2}+4-4a)=24({a}^{2}+2) \ \Rightarrow 9{a}^{2}+36-36a=24{a}^{2}+48 \ \Rightarrow 15{a}^{2}+36a+12=0 \ \Rightarrow 5{a}^{2}+12a+4=0 \ \Rightarrow 5{a}^{2}+10a+2a+4=0 \ \Rightarrow 5a(a+2)+2(a+2)=0 \ \Rightarrow a=\frac{-2}{5},-2 \end{gathered}$

So, $a=-2$ [$\because a\in Z$]

Hence, the eqn. of plane is $-2x+y-z-b=0$

$\begin{gathered} d=|\frac{-12-6-4-b}{\sqrt{4+1+1}}|=3\sqrt{6} \ \Rightarrow |-(b+22)|=18 \ \Rightarrow b=18-22=-4 \ 	herefore {a}^{4}+{b}^{2}=(-2{)}^{4}+(-4{)}^{2} \ =16+16=32 \end{gathered}$

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