Parameterization of Lines in Space

A line inside a three-dimentional coordinate system

Lines in three-dimensional space can be described by using parametrization. The parametric equation of a line through (x0,y0,z0) in the direction of the vector v = (a,b,c) is written like this:

Theory

Parametrization of a Line on Coordinate Form

x = x0 + at y = y0 + bt z = z0 + ct

x = x0 + at y = y0 + bt z = z0 + ct

This can also expressed as a vector in this way:

Theory

Parametrization of a Line on Vector Form

(x,y,z) = (x0,y0,z0) + t (a,b,c) = (x0 + ta,y0 + tb,z0 + tc)

Example 1

Find a parametric equation for the line through the point P = (0, 0, 2) in the direction of v = (2,1, 1).

You set up the expression on vector form, which gives you

(x,y,z) = OP + tv = (0, 0, 2) + t (2,1, 1) = (2t,t, 2 + t).

This can be written on coordinate form like this:

x = 2t y = t z = 2 + t

In three dimensions, we need two equations to describe a line. They’re written like this:

x x0 a = y y0 b = z z0 c

This is another way of writing the two equations

x x0 a = y y0 b

and

y y0 b = z z0 c .

When you have a parametric equation of a line, you can find these equations by rearranging the coordinate form of the parametric equation to all be expressions for t. Then you can set these expressions equal to each other to get the equations above.

Example 2

Find the equation for the line with the parametric equation

x = 2t y = t z = 2 + t.

You rearrange the three expressions to get t alone in all of them. That will give you

t = x 2, t = y = y 1, t = z 2 = z 2 1 .

That makes the equations for the line

x 2 = y 1 = z 2 1

If either a, b or c in the parametric equation is equal to 0, we can’t write the equations in this way. In that case, one of the expressions in the parametric equation will show that one of the variables is constant. That then becomes an equation on its own, and the equation above will consist of just the two remaining expressions instead.

Example 3

Find the equation for the line with this parametric equation

x = 3 y = 3t 4 z = 2t + 1.

You can’t solve the x-equation for t, so you leave it as it is. You solve the others for t and get

t = y + 4 3 , t = z 1 2 .

Thus, the equations for the line are

x = 3 y + 4 3 = z 1 2 .

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