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

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

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

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