Intercept Form of the Equation of a Plane

Suppose the plane $\vec r\cdot\vec n=q$ meets the coordinate axes at $A,B,C$ with intercepts $OA=a,\ OB=b,\ OC=c$. Since $A=(a,0,0)$ lies on the plane, $a\hat i\cdot\vec n=q$, i.e. $\hat i\cdot\vec n=q/a$; similarly $\hat j\cdot\vec n=q/b$ and $\hat k\cdot\vec n=q/c$. Substituting $\vec r=x\hat i+y\hat j+z\hat k$ into $\vec r\cdot\vec n=q$ gives $x(\hat i\cdot\vec n)+y(\hat j\cdot\vec n)+z(\hat k\cdot\vec n)=q$, i.e. $\dfrac{qx}{a}+\dfrac{qy}{b}+\dfrac{qz}{c}=q$. Dividing throughout by $q$ gives the intercept form:

the equation of the plane cutting intercepts $a,b,c$ on the $x,y,z$-axes respectively.

Theorem 6.16. Every first-degree equation $ax+by+cz+d=0$ in $x,y,z$ represents a plane. Proof. Rewrite it as $(x\hat i+y\hat j+z\hat k)\cdot(a\hat i+b\hat j+c\hat k)=-d$, i.e. $\vec r\cdot\vec n=-d$ — the standard vector-form equation of a plane, with $\vec n=a\hat i+b\hat j+c\hat k$ as its normal. …