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Bipin Dubey (13679)

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Equation of plane passing through (x₁,y₁,z₁) be
a(x-x₁) + b(y-y₁) + c(z-z₁) = 0

Since (x₂,y₂,z₂) passes through it
a(x₂-x₁) + b(y₂-y₁) + c(z₂-z₁) = 0

Also the plane is prependicular to a given plane a₁x+b₁y+c₁z = d
Hence aa₁ + bb₁ + cc₁ = 0

Solving these two we get

a=k[(y₂-y₁)c₁-(z₂-z₁)b₁]

b=k[(z₂-z₁)a₁-(x₂-x₁)c₁]

c=k[(x₂-x₁)b₁-(y₂-y₁)a₁]

Substitute this in a(x-x₁) + b(y-y₁) + c(z-z₁) = 0 and cancel out k to obtain required equation of plane.

Bipin Kumar Dubey Chemical Dept. IIT Kharagpur

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