I am having a blank moment.
I have a point and a parametric equation for a line. I need to find the equation of the line through that point that is perpendicular to the line I already have.
the point is (3,1,-2) and the line is x=-1, y = -2+t, z = -1+t
I know the dot product will be zero meaning the are orthogonal and I know it has something to do with find an intersection point but drawing a blank on how Ifind the intersection point that gives me the proper coordinates to have orthogonal lines
"Perpendicular" in three dimensions is different from perpendicular in two dimensions. In two dimensions, perpendicular lines intersect. In three dimensions, they don't need to. Hence, there are an infinite number of lines that are perpendicular to the given line and pass through the given point. There is either exactly one line or an infinite number of lines perpendicular to a given line that passes through BOTH that line AND a given point. For this particular given line and point, there is exactly one. I assume you want the equation for that line.
Since the line passes through both the given point and the given line, a vector in the direction of this perpendicular line would be the vector $\displaystyle \begin{pmatrix}3-(-1) \\ 1-(-2+t) \\ -2-(-1+t)\end{pmatrix} = \begin{pmatrix}4 \\ 3-t \\ -1-t\end{pmatrix}$.
Let's find a vector in the direction of the given line by finding two points along the line (we can choose any two points, so let's say $\displaystyle t=0, t=1$). So, a vector in the direction of the given line is $\displaystyle \begin{pmatrix}-1-(-1) \\ -2+1 - (-2+0) \\ -1+1 - (-1+0)\end{pmatrix} = \begin{pmatrix}0 \\ 1 \\ 1\end{pmatrix}$. Since we want these vectors to be perpendicular, we want the dot product to be zero. Hence:
$\displaystyle (4\cdot 0) + (1\cdot (3-t)) + (1\cdot (-1-t)) = 2-2t = 0$ implies $\displaystyle t=1$. Hence, a vector in the direction of the perpendicular line is $\displaystyle \begin{pmatrix}4 \\ 3-1 \\ -1-1\end{pmatrix} = \begin{pmatrix}4 \\ 2 \\ -2\end{pmatrix}$.
So, the perpendicular line would have parametric equation $\displaystyle x=3+4t, y=1+2t, z=-2-2t$.
so the point (does this mean you are treating the point as a bounded vector) minus the line gives a direction vector along the perpendicular line then I take the given direction vector of the line which is the same you got setting t values <0,1,1>
then set up the dot product and solve for t and input t into the direction vector I got from taking the point minus the line, then write the new parametric equations based off my original point and the direction vector
I'll do that way
but for sake of completeness
(1) the book suggests find the intersection
(2) can I use projection to find it as well??
I got two points along the given line by plugging in $\displaystyle t=0$ and $\displaystyle t=1$. When $\displaystyle t=0$, I get the point $\displaystyle (-1,-2+0,-1+0) = (-1,-2,-1)$. When $\displaystyle t=1$, I get the point $\displaystyle (-1,-2+1,-1+1) = (-1,-1,0)$. So, the vector from the first point to the second point is $\displaystyle \begin{pmatrix}-1-(-1) \\ -1-(-2) \\ 0-(-1)\end{pmatrix} = \begin{pmatrix}0 \\ 1 \\ 1\end{pmatrix}$. That is a vector in the direction of the given line (Essentially, this a vector containing some multiple of the coefficients of $\displaystyle t$ in the parametric equation of the given line). You can rewrite the line as $\displaystyle \begin{pmatrix}-1 \\ -2 \\ -1\end{pmatrix} + t\begin{pmatrix}0 \\ 1 \\ 1\end{pmatrix}$. This immediately gives a vector in the direction of the given line to be $\displaystyle \begin{pmatrix}0 \\ 1 \\ 1\end{pmatrix}$.
Essentially, what I did was I found all possible lines that intersect both the given line and pass through the point $\displaystyle (3,1,-2)$. The point of intersection (in reference to your question 1) would be some point on the line, so it intersects your given line at $\displaystyle (-1,-2+t,-1+t)$ for some value of $\displaystyle t$. We are looking to find what value of $\displaystyle t$ will make the line perpendicular. That's pretty much all I did.
For (2), I am not sure what you mean. What projection are you trying to use to find it?
orthogonal projection:
$\displaystyle \vec{p} = \frac{\vec{a} * \vec{v}}{||a||^2}\vec{a}$
'a' being the direction vector on the given line, 'v' being a vector from a point on the line to the point then from the tip of v to the tip of p there is an orthogonal line