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3 votes
find the equation of the plane which passes through $(2,4,-1)$ and having the normal vector $(2,3,4).$

1 Answer

5 votes

Answer: 2x+3y+4z = 12

Other answers are possible.

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Step-by-step explanation:

The normal vector is n = (a,b,c) = (2,3,4)

The point on the plane is (p,q,r) = (2,4,-1)

Plug those values into the template below and simplify

a(x-p) + b(y-q) + c(z-r) = 0

2(x-2) + 3(y-4) + 4(z-(-1)) = 0

2(x-2) + 3(y-4) + 4(z+1) = 0

2x-4 + 3y-12 + 4z+4 = 0

2x+3y+4z-12 = 0

2x+3y+4z = 12

This is one way to express the cartesian form of the equation.

answered
User Nils
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