Lines and Planes in Space

Directions define lines; normals define planes

A line through $r_0$ with nonzero direction $v$ has parametrization $r(t)=r_0+tv$ . A plane through $p$ with nonzero normal $n$ satisfies $n\cdot(r-p)=0$ . Direction vectors lie along a line, whereas a plane normal is perpendicular to every direction in the plane.

A plane equation $ax+by+cz=d$ has normal $(a,b,c)$ . Scaling every coefficient by the same nonzero number leaves the plane unchanged. A line can lie in a plane, miss it while parallel, or meet it once; substitution distinguishes these cases.

Visual guide

VISUAL GUIDEIntersect a line with a plane

Line

This oblique projection shows r(t) = (1, 0, 2) + t(1, 2, −1) crossing the plane x + y + z = 6 at t = 1.5, the point (2.5, 3, 0.5). Projection preserves incidence but does not show true 3D lengths or angles.

Worked example: intersect a line and a plane

Take $r(t)=(1,0,2)+t(1,2,-1)$ and plane $x+y+z=6$ . Substitution gives $(1+t)+2t+(2-t)=6$ , so $t=3/2$ . The intersection is $(5/2,3,1/2)$ .

Here $n\cdot v=(1,1,1)\cdot(1,2,-1)=2\ne0$ , ensuring exactly one intersection. If this dot product were zero, the line would be parallel to the plane; checking its starting point would tell whether the whole line lies in it.

Worked example: distance from a point

The perpendicular distance from $q$ to $n\cdot r=d$ is

\operatorname{dist}(q,\Pi)=\frac{|n\cdot q-d|}{\|n\|}.

For $q=(0,0,0)$ and $x+2y+2z=9$ , the distance is $9/3=3$ . Dividing by the normal's length makes the answer independent of how the plane equation is scaled. The signed expression without the absolute value measures oriented distance along the chosen normal.

Two nonparallel planes intersect in a line whose direction is the cross product of their normals. Two lines in space, however, can be skew: nonparallel and nonintersecting. A flat drawing can hide their separation in the third coordinate.

Practice

Parametrize the line through $(1,2,3)$ parallel to $(0,1,-1)$ .
Find a plane through $(1,0,0)$ normal to $(2,-1,3)$ .
Do lines $(t,0,0)$ and $(0,s,1)$ intersect?

Show worked solutions

$(x,y,z)=(1,2+t,3-t)$ .
$2(x-1)-y+3z=0$ , or $2x-y+3z=2$ .
No. Their $z$ coordinates are always $0$ and $1$ . Their directions are nonparallel, so they are skew.

Further study

MIT OpenCourseWare: Multivariable Calculus provides a full university course with additional lectures and exercises.