**Case 1** - Finding the intersection point is straight forward if you have Cartesian equations for both lines. If you do not have Cartesian equations for both lines it might be worth finding them…

With two Cartesian equations you simply find the intersection of the two lines as you would in a basic Algebra course. Use your GDC or solve one equation for *y* (or *x*) and substitute into the second equation. Voila!

**Case 2** - 3D vectors… Each line will have its own "time" variable. Often represented by *t* and *s* when there are two vector equations. If the two lines intersect then there is a value for *t* and a value for *s* that will results in the same *(x,y,z)* coordinates for both lines. Essentially we set the two equations equal to each other and solve for *t* and *s*.

#### Example of Find an Intersection Point

Given the two vector equations $\binom{x}{y}=\binom{1}{3}+t \binom{3}{-1}$ and $\binom{x}{y}=\binom{2}{2}+t \binom{-1}{3}$ find the point $(x,y)$ where they intersect.

First we find the Cartesian equations. We found the first above to be:

(11)
\begin{equation} 10-x-3y=0 \end{equation}

Similarly we can find a Cartesian equation for the second:

(12)
\begin{equation} x=2-t \end{equation}

(13)
\begin{equation} y=2+3t \end{equation}

Solving for *t* and setting the two equation equal:

(14)
\begin{align} 2-x=\frac{y-2}{3} \end{align}

Simplifying:

(15)
\begin{equation} 8-3x-y = 0 \end{equation}

Solving the second Cartesian equation for *y* ($y=8-3x$) and substituting into the first equation we get:

(16)
\begin{equation} 10-x-3(8-3x)=-14 +8x = 0 \end{equation}

Resulting in $x=\frac{14}{8}=\frac{7}{4}=1.75$ and $y= 2.75$. We can confirm this graphically as shown below.

#### Example of Finding Intersections in 3D

Given the two vector equations

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\begin{align} r_1 = {\begin{pmatrix} 1\\ -1\\ 4 \end{pmatrix}} + t\begin{pmatrix} 1\\ -1\\ 1 \end{pmatrix} \end{align}

and

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\begin{align} r_2 = {\begin{pmatrix} 2\\ 4\\ 7 \end{pmatrix}} + s{\begin{pmatrix} 2\\ 1\\ 3 \end{pmatrix}} \end{align}

Since we only have two unknowns to solve for *t* and *s* we only need two equations. Lets take the first two rows of each equation:

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\begin{matrix} x=1+t\\ y=-1-t \end{matrix}

(20)
\begin{matrix} x=2+2s\\ y=4+s \end{matrix}

Each of the x-equations can be set equal and each of the y-equations set equal. This results in two equations and two unknowns. Solve… We find that $t=-3, s=-2$. One of these values is then substituted back into the corresponding original equation to solve for the *(x,y,z)* coordinates. Substituting $t=-3$ into the first equation:

(21)
\begin{align} r_1 = \begin{pmatrix} 1\\ -1\\ 4 \end{pmatrix} + (-3){\begin{pmatrix} 1\\ -1\\ 1 \end{pmatrix}} = {\begin{pmatrix} -2\\ 2\\ 1 \end{pmatrix}} \end{align}

You can also check your work by substituting $s=-2$ into the other equation and if all is well you will get the same coordinate point for the intersection.

do you have the 3d vectors notes>

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