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Equation of the plane containing the straight line `x/2=y/3=z/4` and perpendicular to the plane containing the straight lines `x/2=y/4=z/2` and `x/4=y/2=z/3` is

A

`x+2y-2z=0`

B

`3x+2y-2z=0`

C

`x-2y+z=0`

D

`5x+2y-4z=0`

Text Solution

Verified by Experts

The correct Answer is:
c
Plane `1:ax+by+cz=0` contains line `(x)/(2)= (y)/(3)= (z)/(4)`
`therefore" "2a+3b+4c=0`
Plane `2 : a'x+b'y+c'z=0` is perpendicular to plane containing lines `(x)/(3)= (y)/(4)= (z)/(2) and (x)/(4)= (y)/(2) = (z)/(3)`
`therefore " "3a'+4b' + 2c'=0 and 4a'+ 2b' + 3c' =0`
`rArr" "(a')/(12-4)= (b')/(8-9)= (c')/(6-16)`
`rArr" "8a-b-10c =0" "`(ii)
From (i) and (ii),
`" "(a)/(-30+4)= (b)/(32+20)= (c)/(-2-24) `
`rArr" "` Equation of plane `x-2y+z=0`
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