Mathematics · Ch 6 — Applications of Vector Algebra

Equation of a Plane Perpendicular to a Vector and Passing Through a Given Point

6.8.2

Vector form. Let the plane pass through a point $A$ (position vector $\vec a$ ) with normal vector $\vec n$ , and let $P$ (position vector $\vec r$ ) be any point of the plane. Then $\vec{AP}$ is perpendicular to $\vec n$ , so $\vec{AP}\cdot\vec n=0$ , i.e. $(\vec r-\vec a)\cdot\vec n=0.$
Note

This can be rearranged as $\vec r\cdot\vec n=\vec a\cdot\vec n$ , i.e. $\vec r\cdot\vec n=q$ where $q=\vec a\cdot\vec n$ — the same standard form seen in §6.8.1.
Cartesian form. With $\vec n=a\hat i+b\hat j+c\hat k$ (direction ratios $a,b,c$ ) and $A=(x_1,y_1,z_1)$ , substituting and dotting with $\hat i,\hat j,\hat k$ gives $a(x-x_1)+b(y-y_1)+c(z-z_1)=0.$ …