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Question

Find the equation of the plane passing through the point (2,1,1) and through the line of intersection of the planes r(2^i3^j+^k)=3 and r(^i+5^j^k)=4.4.

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Solution

We know that equation of plane passing through the line of intersection of the planes is
r.(a+λb)=c1+λc2
where a and b are the normal vectors of the planes and c1 and c2 are the intercept terms of the respective planes.

So, here, equation of the required plane is,
r.((2+λ)^i+(3+5λ)^j+(1λ)^k)=3+4.4λ

It is given that (2,1,1) passes through the plane.

(2^i^j+^k).((2+λ)^i+(3+5λ)^j+(1λ)^k)=3+4.4λ
2(2+λ)(3+5λ)+(1λ)=3+4.4λ
4+2λ+35λ+1λ=3+4.4λ
8.4λ=5

Substituting λ=5/8.4 we get the equation as,
r.((21.8)^i+(0.2)^j+(3.4)^k)=47.2


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