Science:Math Exam Resources/Courses/MATH105/April 2017/Question 01 (a)/Solution 1

If $\langle x,y,z\rangle$ is some point on the plane, then it forms a vector ${\mathbf {v}}=(\langle x,y,z\rangle -\langle 0,-3,2\rangle )$ with the given point on the plane $\langle 0,-3,2\rangle$. Since a normal vector on a plane is orthogonal to any vector lying on the plane, this vector ${\mathbf {v}}$ is therefore perpendicular to the normal vector $\langle 2,-1,4\rangle$ of the plane.

This implies that $(\langle x,y,z\rangle -\langle 0,-3,2\rangle )\cdot \langle 2,-1,4\rangle =0$.

Thus the equation of the plane is $\color {blue}2x-y+4z=11$.