
# 12.5: Lines and Planes

Lines and planes are perhaps the simplest of curves and surfaces in three dimensional space. They also will prove important as we seek to understand more complicated curves and surfaces.

The equation of a line in two dimensions is $$ax+by=c$$; it is reasonable to expect that a line in three dimensions is given by $$ax + by +cz = d$$; reasonable, but wrong---it turns out that this is the equation of a plane.

A plane does not have an obvious "direction'' as does a line. It is possible to associate a plane with a direction in a very useful way, however: there are exactly two directions perpendicular to a plane. Any vector with one of these two directions is called normal to the plane. So while there are many normal vectors to a given plane, they are all parallel or anti-parallel to each other.

Suppose two points $$(v_1,v_2,v_3)$$ and $$(w_1,w_2,w_3)$$ are in a plane; then the vector $$\langle w_1-v_1,w_2-v_2,w_3-v_3\rangle$$ is parallel to the plane; in particular, if this vector is placed with its tail at $$(v_1,v_2,v_3)$$ then its head is at $$(w_1,w_2,w_3)$$ and it lies in the plane. As a result, any vector perpendicular to the plane is perpendicular to $$\langle w_1-v_1,w_2-v_2,w_3-v_3\rangle$$. In fact, it is easy to see that the plane consists of precisely those points $$(w_1,w_2,w_3)$$ for which $$\langle w_1-v_1,w_2-v_2,w_3-v_3\rangle$$ is perpendicular to a normal to the plane, as indicated in figure 12.5.1.

Turning this around, suppose we know that $$\langle a,b,c\rangle$$ is normal to a plane containing the point $$(v_1,v_2,v_3)$$. Then $$(x,y,z)$$ is in the plane if and only if $$\langle a,b,c\rangle$$ is perpendicular to $$\langle x-v_1,y-v_2,z-v_3\rangle$$. In turn, we know that this is true precisely when $$\langle a,b,c\rangle\cdot\langle x-v_1,y-v_2,z-v_3\rangle=0$$. That is, $$(x,y,z)$$ is in the plane if and only if

\eqalign{ \langle a,b,c\rangle\cdot\langle x-v_1,y-v_2,z-v_3\rangle&=0\cr a(x-v_1)+b(y-v_2)+c(z-v_3)&=0\cr ax+by+cz-av_1-bv_2-cv_3&=0\cr ax+by+cz&=av_1+bv_2+cv_3.\cr }

Working backwards, note that if $$(x,y,z)$$ is a point satisfying $$ax+by+cz=d$$ then

\eqalign{ ax+by+cz&=d\cr ax+by+cz-d&=0\cr a(x-d/a)+b(y-0)+c(z-0)&=0\cr \langle a,b,c\rangle\cdot\langle x-d/a,y,z\rangle&=0.\cr }

Namely, $$\langle a,b,c\rangle$$ is perpendicular to the vector with tail at $$(d/a,0,0)$$ and head at $$(x,y,z)$$. This means that the points $$(x,y,z)$$ that satisfy the equation $$ax+by+cz=d$$ form a plane perpendicular to $$\langle a,b,c\rangle$$. (This doesn't work if $$a=0$$, but in that case we can use $$b$$ or $$c$$ in the role of $$a$$. That is, either $$a(x-0)+b(y-d/b)+c(z-0)=0$$ or $$a(x-0)+b(y-0)+c(z-d/c)=0$$.)

### Contributors

• Integrated by Justin Marshall.