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If the plane \mathrm{2 x+y-5 z=0} is rotated about its line of intersection with the plane \mathrm{3 x-y+4 z-7=0} by an angle of \mathrm{\frac{\pi}{2}}, then the plane after the rotation passes through the point :

Option: 1

(2,-2,0)


Option: 2

(-2,2,0)


Option: 3

(1,0,2)


Option: 4

(-1,0,-2)


Answers (1)

best_answer

\begin{aligned} &(2 x+y-5 z)+\lambda(3 x-y+4 z-7)=0 \\ &(2+3 \lambda) x+(1-\lambda) y+(-5+4 \lambda) z-7 \lambda=0 \\ & \text{For this plane to be perpendicular to 2x+y-5z=0} \\ & 2(2+3 \lambda)+(1-\lambda)-5(-5+4 \lambda)=0 \\ &-15 \lambda=-30 \\ &\lambda=2 \end{aligned} \\ \therefore \text{Required plane}: 8 x-y+3 z=14

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Kuldeep Maurya

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