Suppose there is a light source at $(0, 0, 8)$. Consider the line segment $r(t) = \langle 3-3t, 1+t, 2+2t\rangle, 0\le t\le1$. Find the shadow cast by the line segment on the plane $x+y-2z=8$
I have never encountered this type of problem before, but nevertheless I gave it a try. I would imagine that we would first find the plane that the point's shadow draws. This seems to be the plane passing through $(3, 1, 2)$, $(0, 2, 4)$, and $(0, 0, 8)$. This gives me the plane equation $4x+6y+3z=24$.
The intersection of the two planes is $(x, y, z) = (13t+16, -11t-8, t)$
But now I don't know how to account for the line segment's ends.
How would I continue, and is there a better method?
Thanks
Based on the comment, we can do as follows.
Since the line segment ends at $(3, 1, 2)$ and $(0, 2, 4)$, we can construct two lines from the light source, which will be the edges of our "shadow".
These, namely, are $$\text{line1}(t) = \langle 3t, t, 8-6t \rangle$$ $$\text{line2}(t) = \langle 0, 2t, 8-4t \rangle$$
We solve for the intersection of the plane by substituting the entire line into the plane equation and solving for $t$, which give us $t=\frac32\Longleftrightarrow (\frac92, \frac32, -1)$ for line $1$ and $t=\frac{12}5 \Longleftrightarrow (0, \frac{24}5, -\frac85)$ for line $2$.
Parameterizing this gives $$\text{shadow}(t) = \left\langle \frac92 - \frac92 t, \frac32 + \frac{33}{10} t, -1-\frac35 t\right\rangle$$ for $0<= t<= 1$