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Find the equation of the plane through t...

Find the equation of the plane through the line of intersection of the planes x+y+z=1 and 2x+3y+4z=5 which is perpendicular to the plane x-y+z=0

A

`r.(hati-hatk)-2=0`

B

`rxx(hati+hatk)+2=0`

C

`rxx(hati-hatk)+2=0`

D

`r.(hati-hatk)+2=0`

Text Solution

Verified by Experts

The correct Answer is:
D
Since, equation of planes passes through the line of intersection of the planes
`x+y+z=1`
and `2x+3y+4z=5,` is
`(x+y+z-1)+lambda(2x+3y+4z-5)=0`
`implies" "(1+2lambda)x+(1+3lambda)y+(1+4lambda)z-(1+5lambda)=0" "...(i)`
`because` The plane (i) is perpendicular to the plane
`x-y+z=0`.
`:.(1+2lambda)-(1+3lambda)+(1+4lambda)=0`
[`because` if plane `a_(1)x+b_(1)y+c_(1)z+d_(1)=0` is perpendicular to plane `a_(2)x+b_(2)y+c_(2)z+d_(2)=0," then"a_(1)a_(2)+b_(1)b_(2)+c_(1)c_(2)=0`]
`implies" "3lambda+1=0`
`implies" "lambda=-(1)/(3)" "...(ii)`
So, the equation of required plane, is
`(1-(2)/(3))x+(1-(3)/(3))y+(1-(4)/(3))z-(1-(5)/(3))=0`
`implies(1)/(3)x-(1)/(3)z+(2)/(3)=0impliesx-z+2=0`
Now, vector form is `r.(hati-hatk)+2=0`
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