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.
Best Answer
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:
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.
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
$$ x+y=5 $$