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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 xy+z=3

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Solution

Equation of plane passing through intersections of the planes P1:x+y+z1=0

and P2:2x+3y+4z5=0 is given by,

P2+λP1=0

(2+λ)x+(3+λ)y+(4+λ)z(5+λ)=0.... (1)

Now for above plane to be perpendicular to the given plane xy+z=0

Dot product of their normal vector should be zero

(2+λ)(1)+(3+λ)(1)+(4+λ)1=0

3+λ=0λ=3

Hence the required plane is,

(23)x+(33)y+(43)z(53)=0, substitute λ=3 in (1)

x+z2=0

xz+2=0

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