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I am attempting to do this problem for my calc homework and I am stuck on trying to set up the equations to find the intersection point. I understand that the equations that are given to me are in the form of a vector equation, $v = r + tv$ where $r$ and $v$ are vectors. I'm not exactly sure how to get the points and the equation from these vector equations tho.

Consider the lines below.

$r = \langle 2, 3, 0\rangle + t \langle3, -3, 3\rangle$

$r = \langle5, 0, 3\rangle + s\langle-3, 3, 0\rangle$

(a) Find the point at which the given lines intersect. $(x= ,y= ,z= )$

(b) Find an equation of the plane that contains these lines.

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1 Answer

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Set the two equations equal to each other. Then for each of the three coordinates, you will get an equation in $s$ and $t$, so you will have three equations in two variables. If the two lines are co-planer then there is a unique solution for $s$ and $t$ which will give you the coordinates of the point of intersection.

For example, the first coordinates give you the equation

$$ 2+3t=5-3s $$

Find the equations for the other two coordinates and finish the problem.

ADDENDUM:

Now that you correctly found the point of intersection $(5,0,3)$ you have the necessary information to find the equation of the plane which contains the two intersecting lines.

To find the equation of a plane containing two intersecting lines you need three pieces of information: direction vectors for each of the two lines and the point of intersection of the two lines.

The direction vectors are the vector coefficients of your two vector line equations:

1. $\langle 3,-3,3\rangle$
2. $\langle 3,-3,0\rangle$

These two may be simplified by multiplying by $\dfrac{1}{3}$ since multiplication by a nonzero constant does not change the direction of a vector. So use the following for the two direction vectors.

1. $\langle 1,-1,1\rangle$
2. $\langle 1,-1,0\rangle$

The cross-product of these two vectors gives a normal vector $N=\langle a,b,c\rangle$ for the plane containing the two lines.

$N=\begin{vmatrix} \mathbf{i}&\phantom{-}\mathbf{j}&\mathbf{k}\\1&-1&1\\1&-1&0\end{vmatrix}=\mathbf{i}+\mathbf{j}=\langle 1,1,0\rangle$

The equation of a plane has the form

$$ ax+by+cz=d $$

where the normal vector is $\langle a,b,c\rangle$ and $d$ is computed using the values of $a,b,c$ and the coordinates of a point in the plane ( in this case, $(5,0,3)$).

So the equation for this plane is

$$ 1\cdot x+1\cdot y+0\cdot z=d$$

where

$$1(5)+1(0)+0(3)=d=5$$

So the equation of the line containing the two lines is

$$ x+y=5 $$

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