Chapter 13

Section 13.5Lines

Equations of LinesTo form the equation of a line in 3 dimensions we need two pieces of information. A point on the line and a vector d called the direction vector that tells us the orientation (“slope“) of the line. Once you are on the line to get to another point it is just a scalar multiple of d (i.e. ). The location of another point

x

y

z𝑙 (𝑡 )

(𝑥0 , 𝑦0 , 𝑧0 )(𝑥 (𝑡 ) , 𝑦 (𝑡 ) ,𝑧 (𝑡 ) )

𝑡 d

Vector Form

Parametric Form

ExampleFind the vector equation passing through the point and parallel to the line with equation given to the right.

𝑙 (𝑡 )= ⟨7−4 𝑡 ,8,3 𝑡−1 ⟩

We already know the point is on the line so we need to find the direction d. Since the lines are parallel they will have parallel direction vectors. The coefficients of t form the direction vector d.

ExampleFind the vector equation of the line passing through the points and .

A direction vector d for the line can be found by finding the vector from the first point to the second.

ExampleFind the equation of the line that is perpendicular to both of the vectors and passing through the point .

To get a direction vector d that is perpendicular to both vectors u and v we can use the cross product.

To get the equation for the line we plug the vector d and the point into the equation.

ExampleDetermine if the lines and given to the right intersect. If they intersect find the point of intersection.

The tricky part here is to realize that the value for t that gives the point of intersection in might not be the same value of t that gives the point of intersection in . Choose a different independent variable for , say u.

This gives a system of equations with the variables t and u. Solve this system for t and u. This can be done in many ways here we equate the x and y components.

{ 2 𝑡=−𝑢+1𝑡−5=2𝑢+3

={𝑢+2𝑡=1𝑡=2𝑢+8

={𝑢+2 (2𝑢+8 )=1𝑡=2𝑢+8

={5𝑢=−15𝑡=2𝑢+8

={𝑢=−3𝑡=2

Substituting the values for t and u into their equations we see the lines intersect at the point .

Symmetric Equation of a LineBeside the vector form of a line and the parametric form of a line, a line can also be expressed as three equivalent expressions of the x, y and z variables. This is done by solving the parametric system for t and equating them.

Direction vector Point

or equivalently,

ExampleA line l is parallel to the line with symmetric equations given to the right and passes through the point . Find the vector form of the line l.

3𝑥−76

=−2 𝑦+84

=𝑧5

In order to determine the direction vector d the coefficients of x, y and z all need to be 1. Divide the first expression by 3 and the second by -2.

𝑥−732

=𝑦−4−2

=𝑧5

This gives that which gives parallel line