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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=106`

B

`x-8y+13z=103`

C

`x-4y+6z=110`

D

`x-8y+13z=19=105`

Text Solution

Verified by Experts

The correct Answer is:
a
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)+ lamda(5x+ 3y+ 10z- 25)=0`
or `" "(4+5lamda)x +(7+ 3lamda ) y + (4+ 10 lamda)z+ 81- 25 lamda =0 " "` (i)
which is perpendicular to `4x+ 7y + 4z+ 81=0`
`rArr" "4(4+ 5lamda)+7(7+ 3lamda) + 4(4+ 10 lamda) =0`
or `" " 81 lamda + 81=0`
or `" "lamda=-1`
Hence, the plane is `x-4y+6z= 106`
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