Planes in 3D Math Example 1

Follow the full solution, then compare it with the other examples linked below.

Example 1

easy

Write the equation of the plane with normal vector

\langle 2, -3, 1 \rangle

passing through the point

(1, 4, -2)

Solution

1
The equation of a plane with normal $\langle A, B, C \rangle$ through $(x_0, y_0, z_0)$ is $A(x-x_0) + B(y-y_0) + C(z-z_0) = 0$ .
2
Substitute: $2(x-1) - 3(y-4) + 1(z+2) = 0$ .
3
Expand: $2x - 2 - 3y + 12 + z + 2 = 0$ , so $2x - 3y + z + 12 = 0$ .

Answer

2x - 3y + z + 12 = 0

A plane in 3D is determined by a point on the plane and a normal vector perpendicular to it. The normal vector's components become the coefficients of