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Question
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. Also find the distance of the plane, obtained above, from the origin.
Solution
Equation of the plane through the line of intersection of the two planes is
` (x+y+z−1)+λ(2x+3y+4z−5)=0⇒(1+2λ)x+(1+3λ)y+(1+4λ)z−1−5λ=0 .........(1)`
Normal vector of the required plane is
`vecN =(1+2λ)hati+(1+3λ)hatj+(1+4λ)hatk`
Since this plane is perpendicular to x-y+z=0, their normals are also perpendicular.
`⇒vecN_1.vecN_2=0`
`⇒((1+2λ)hati+(1+3λ)hatj+(1+4λ)hatk).(hati−hatj+hatk)=0`
`⇒1+2λ−1−3λ+1+4λ=0`
`⇒λ=−1/3`
Substituting the value of λ in equation (1), to obtain the required equation of the plane
`(1+2(−1/3))x+(1+3(−1/3))y+(1+4(−1/3))z−1−5(−1/3)`
=0
⇒x−z+2=0
To find out the distance of this plane from the origin we need to convert this equation to the standard form:
lx+my+nz=d
Dividing both sides of the equation by `sqrt2`
, we get:
`x/sqrt2−z/sqrt2=−2/sqrt2`
`x/sqrt2−z/sqrt2=−sqrt2`
Hence, the distance of the plane from the origin is `sqrt2` units.
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