Finding the equation of a plane, provided a line and a point?

Question

Question: Given the line $$\begin{pmatrix} x \\ y \\ z \\ \end{pmatrix} = \begin{pmatrix} 1 \\ -3 \\ 2 \\ \end{pmatrix} + t \begin{pmatrix} -2 \\ 4 \\ 7 \\ \end{pmatrix},$$ find a plane which is intersected by the line at the point A[3, -7, -5].

So far, I've found another point on the plane, B[-1,1,9], by subbing 1=t, and found the direction vector of the vector (u=AB) that ended up being [-4,8,14]. I'm currently stuck with finding the normal vector--I keep getting [0,0,0] as an answer. I noticed that it might be because the direction vector found is a scalar multiple of the one of the line, but this is just an assumption.

Does anyone know where I can go from here?

Answer 1

The easiest way is looking for parametric equation of this plane: you have point $A$, so the parametric equation is $[3,-7,-5]+t\overline{v}+s\overline{w}$, where $t,s \in \mathbb{R}$. As $\overline{v}$ and $\overline{w}$ you can choose linear independent vectors which aren't parallel to $[-2, 4, 7]$ (because line should intersect the plane), for example $[1,0,0]$ and $[0,1,0]$.

Answer 2

There are many possible planes that work here, so why not take the plane whose normal vector is given by the direction vector of the given line? Hence, we know that the plane has the form: $$ -2x + 4y + 7z = D $$ for some constant $D$. Now since this plane must contain the given intersection point, substitution yields: $$ D = -2(3) + 4(-7) + 7(-5) = -6 - 28 - 35 = -69 $$

Multiplying through by $-1$ to make the equation look prettier, we obtain the plane given by: $$ 2x - 4y - 7z = 69 $$