Vector, parametric, and symmetric equations of a line in 3D

Once we know a point on a line and which way the line goes, we have enough information to describe it. We’ll start with the vector equation, then use its components to write the parametric and symmetric equations.

Start with a point and a direction vector

Let P=(x0,y0,z0) be a point on the line, and let v=⟨a,b,c⟩ be a nonzero vector parallel to it. Every point on the line can be reached by adding a real multiple of v to the position vector of P.

r(t)=⟨x0,y0,z0⟩+t⟨a,b,c⟩

Write the vector and parametric equations

Let’s use P=(2,-1,3) and v=⟨2,-1,4⟩. We can plug these directly into the vector equation.

r(t)=⟨2,-1,3⟩+t⟨2,-1,4⟩
One step in the direction of the lineStarting at P, adding the direction vector gives Q. The parameter is zero at P and one at Q.P = (2, −1, 3)Q = (4, −2, 7)v = ⟨2, −1, 4⟩t = 0t = 1
At t = 1, we add one copy of v to the starting point P. The drawing shows the line schematically.

Reading the three components separately gives the parametric equations. All three use the same value of the parameter.

x=2+2ty=-1-tz=3+4t

Eliminate the parameter to write symmetric equations

Solve each equation for t. Because the three expressions all equal the same parameter, we can set them equal to each other.

x-22=y+1-1=z-34

For example, at (4,-2,7) each fraction equals 1, which agrees with the vector equation at t=1.

Keep the coordinate fixed when its direction component is zero

If the direction is ⟨0,1,-2⟩ and we start at the same P, the first coordinate stays at 2. Solve only the other two parametric equations for t.

x=2,y+11=z-3-2

A plane can also give us the direction. If a line is perpendicular to 2x-y+4z=1, the coefficients of the plane give a direction vector ⟨2,-1,4⟩ for the line. We still need a point to tell us where to place it.

For a different way a pair of planes can determine a line, see finding the line where two planes intersect.