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The plane 4x+7y+4z+81=0 is rotated throu...

The plane `4x+7y+4z+81=0` is rotated through a right angle about its line of intersection with the plane `5x+3y+10 z=25.` The equation of the plane in its new position is a. `x-4y+6z=106` b. `x-8y+13 z=103` c. `x-4y+6z=110` d. `x-8y+13 z=105`

A

x-4y+6z=56

B

x-4y+6z=106

C

5x-4y-z=81

D

x+2y-z=1

Text Solution

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The correct Answer is:
B
The equation of the plane through the line of intersection of the planes
4x+7y+4z+81=0 and 5x+3y+10z=25 is (4x+7y+4z+81)+ `lambda` (5x+3y+10z-25)=0
`rArr (4+5lambda)x+(7+3lambda)y + (4+10 lambda)z+81-25lambda=0`…(i)
This plane is perpendicular to the plane 4x+7y+4z+81=0
So, `4(4+5lambda)+7(7+3lambda)+4(4+10 lambda)=0 rArr 81+81lambda =0 rArr lambda =-1`
Then from equation (i), x-4y+6z=106
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