This is a homework question I am having difficulty answering. (It is Calc III, but this problem only concerns vectors).
(A) Find the parametric equations for the line through the point $P =
(4, -1, 4)$ that is perpendicular to the plane $3x + 1y – 5z = 1$. Use
"$t$" as your variable, $t = 0$ should correspond to $P$, and the velocity
vector of the line should be the same as the normal vector to the
plane found directly from its equation.(B) At what point $Q$ does this line intersect the $yz$-plane?
So Here's what I've tried conceptualizing. First we have point $P$, which we can think of as a vector from the origin. We can also create the normal vector from the plane: $n = \langle 3, 1, -5\rangle$. We can pick a point on the plane that satisfies the equation $p_0 = (2,0,1)$. We can now use this information to start forming a triangle we can solve for. One side: $n – p_0 = \langle 1, 1, -6\rangle$. The hypotenuse can be found with $P – p_0 = \langle 2, -1, 3\rangle$. Which leaves us with one remaining side, which will be a vector within the plane which should give us the point in which the line will intersect with the plane, which we can turn into a vector from the origin and add to the unit vector of $n$ to create our equation.
I think finding this remaining side should be how we get all this information, but I don't know if that's right or if that's even the correct approach. Can anyone give me any advice?
Best Answer
If $\mathbf{n}$ is the normal vector to the given plane and $\mathbf{p}$ is the point through which the line is supposed to pass, then the equation of the line will be of the form $\mathbf{r}=\mathbf{p}+t\mathbf{n}$. You already have both those vectors so you don't need anything else.