3.7 - Vector equations of lines

Vector equation of a line

Here $\mathbf r$ is the position vector of a variable point on the line, $\mathbf a$ is the position vector of one fixed point, $\mathbf b$ is a direction vector, and $t$ is a real scalar parameter.

Worked example 1: reading a line equation

Question. A line has equation $\mathbf r=2\mathbf j-\mathbf k+t(2\mathbf i-\mathbf j+3\mathbf k)$. State a point and a direction vector for the line, and find the point with $t=2$.

Plan. Read the fixed vector, then substitute $t=2$ into all three components.

At $t=2$, this is $4\mathbf i+5\mathbf k$.

Answer. Point $(0,2,-1)$, direction $(2,-1,3)$, and the requested point $(4,0,5)$.

Check. Its displacement from the fixed point is $(4,-2,6)=2(2,-1,3)$, as required.

Worked example 2: the same line at a different parameter scale

Question. A second equation is $\mathbf r=2\mathbf j-\mathbf k+u(4\mathbf i-2\mathbf j+6\mathbf k)$. Find $u$ at $(4,0,5)$ and explain why this equation describes the same line as the previous one.

Plan. Match the displacement from the common fixed point to a multiple of the new direction vector.

The new direction vector is $2\mathbf b$, so $u(2\mathbf b)=(2u)\mathbf b$. As $u$ ranges over all real numbers, $t=2u$ also ranges over all real numbers.

Answer. $u=1$; both equations describe the same set of points.

Check. $u=-\tfrac12$ corresponds to $t=-1$, and both give $(-2,3,-4)$.

Watch out

In $\mathbf r=\mathbf a+t\mathbf b$, $\mathbf a$ is a position vector, not generally a direction vector. A zero direction vector gives only one point, not a line.

A line through two points

For distinct points with position vectors $\mathbf a$ and $\mathbf b$,

Not in the formula booklet: learn it.

Worked example 3: a point on the line

Question. Find an equation of the line through $A=(1,0,2)$ and $B=(3,4,0)$. Determine whether $C=(5,8,-2)$ lies on it.

Plan. Form $\overrightarrow{AB}$, simplify it, then compare all coordinates of $C$.

Hence a suitable equation is

For $C$, the first coordinate gives $1+s=5$, so $s=4$. Then $y=2s=8$ and $z=2-s=-2$.

Answer. The equation above describes the line, and $C$ lies on it.

Check. Independently, $\overrightarrow{AC}=(4,8,-4)=2\overrightarrow{AB}$, so the displacement is parallel to the line.

Worked example 4: rejecting a point

Question. Determine whether $D=(5,8,-1)$ lies on the same line through $A=(1,0,2)$ and $B=(3,4,0)$.

Plan. Use the parameter forced by the first coordinate, then test the remaining coordinates.

This gives $y=8$, but $z=2-4=-2\ne-1$.

Answer. $D$ does not lie on the line.

Check. $\overrightarrow{AD}=(4,8,-3)$ is not a scalar multiple of $(1,2,-1)$: its first component would require multiplier $4$, while its last would require $3$.