Find the cartesian equation of the plane through the point(2,5,3) with normal 3i+2j-7k.
The normal is: n̂ = 3î + 2ĵ – 7k̂.
Let P1 and P be two position vectors of points on the plane, where
P1 is the position vector of the given point on the plane, (2, 5, 3), and
P is the position vector of a general point on the plane, (x, y, z).
Let R be a vector contained within the plane such that R̂ = P̂ – P̂1.
Since n̂ is the normal vector to the plane, then n̂•R̂ = 0.
i.e. n̂•(P̂ – P̂1) = 0
or, n̂•P̂ = n̂•P̂1
Expanding,
(3î + 2ĵ – 7k̂)•(xî + yĵ + zk̂) = (3 2 -7)•(2 5 3)
3x + 2y – 7z = 3*2 + 2*5 – 7*3 = 6 + 10 – 21 = -5
Therefore, equation of the plane is: 3x + 2y – 7z + 5 = 0