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Let the plane P: 4 x-y+z=10 be rotated by an angle\frac{\pi}{2} about its line of intersection with the plane x+y-z =4. If \alpha is the distance of the point (2,3,-4) from the new position of the plane P, then 35 \alpha is equal to

Option: 1

90


Option: 2

105


Option: 3

85


Option: 4

126


Answers (1)

best_answer

Let equation in new position is


\begin{aligned} & (4 x-y+z-10)+\lambda(x+y-z-4)=0 \\ & 4(4+\lambda)-1 .(-1+\lambda)+1 \cdot(1-\lambda)=0 \\ & \Rightarrow \lambda=-9 \end{aligned} \\

So equation in new position is

\begin{aligned} & -5 x-10 y+10 z+26=0 \\ & \Rightarrow \alpha=\frac{54}{15} \end{aligned}

Posted by

HARSH KANKARIA

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