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For a, b \epsilon Z and \left |a-b \right |< 10 let the angle between the plane P: a x+y-z=b \text { and the line } 1: x-1=a-y=z +1 \text { be } \cos ^{-1}\left(\frac{1}{3}\right) . If the distance of the point (6, -6, 4) from the plane P is3 \sqrt{6} \text {, then } a^4+b^2 \text { is equal to }

Option: 1

85


Option: 2

48


Option: 3

25


Option: 4

32


Answers (1)

best_answer

$$ \begin{aligned} & \theta=\cos ^{-1} \frac{1}{3} \therefore \sin \theta=\sqrt{1-\frac{1}{9}}=\frac{2 \sqrt{2}}{3} \\ & \sin \theta=\frac{a \cdot 1+1(-1)+(-1) \cdot 1}{\sqrt{a^2+1+1} \cdot \sqrt{3}}=\frac{2 \sqrt{2}}{3} \\ & \Rightarrow\{3(a-2)\}^2=24\left(a^2+2\right) \\ & \Rightarrow 3\left(a^2-4 a+4\right)=8 a^2+16 \\ & \Rightarrow 5 a^2+12 a+4=0 \\ & \Rightarrow 5 a^2+10 a+2 a+4=0 \\ & \therefore a=-2, \frac{-2}{5} \because a \in z \\ & \therefore a=-2 \end{aligned} $$ Distance of $(6,-6,4)$ from $$ \begin{aligned} & -2 x+y-z-b=0 \text { is } 3 \sqrt{6} \\ & \therefore\left|\frac{-12-6-4-b}{\sqrt{4+1+1}}\right|=3 \sqrt{6} \\ & \Rightarrow|b+22|=18 \therefore b=-40,-4 \\ & \because|a-b| \leq 10 \\ & \therefore \mathrm{b}=-4 \\ & \therefore a^4+b^2 \\ & =32 \mathrm{Ans} . \\ & \end{aligned}

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