I am having some difficulty with this concept. I am studying a problem:
Find an equation for the plane that passes through the point $P = (1, 2, 3)$ and contains the line $L$ given by the parametric equation $x(t) = 1 − 3t,\, y(t) = 3$, and $z(t) = 6 + 2t$.
The solution states that $v = \left< 3, 0, −2 \right>$ is parallel to $L$ and hence the plane. $Q = (1, 3, 6)$ is in the plane and hence $PQ = \left< 0, 1, 3 \right>$ is parallel to the plane. $n = v \times PQ = \left< −2, 9, −3 \right>$ is perpendicular to the plane. Therefore, the plane is $−2x + 9y − 3z = −2 \cdot 1 + 9 \cdot 2 − 3 \cdot 3 = 7$.
Could someone help me with this reasoning? I understand $v$ is parallel to $L$ because it is a scaled version of the normal to $L$, where the normal is $\left< -3, 0, 2 \right>$. I understand why $Q$ is in the plane.
Why is $PQ$ parallel to the plane? Why do we know that $v \times PQ$ will be perpendicular to the plane? Is this because the cross product of two vectors parallel to a plane will be perpendicular to the plane? Thank you for any clarification or help in advance! Just trying to get a better hold on this concept.
To begin with, $v$ isn’t normal to $L$. That’s pretty much the opposite of being parallel. In vector form, the given parameterization of the line is $\langle1,3,6\rangle+t\langle-3,0,2\rangle$ from which it should be obvious that $Q=(1,3,6)$ is a point on the line (for $t=0$) and $v=\langle-3,0,2\rangle$ is the line’s direction vector, so is by definition parallel to $L$.
The points $P$ and $Q$ are on the plane, so the line segment $\overline{PQ}$ is also contained in the plane. The vector $Q-P$ (what you’re calling $PQ$) is parallel to this line segment, which means that it is also parallel to any plane that contains the line segment.
Finally, the cross product of two vectors is perpendicular to them both. Since both of the vectors are parallel to the plane you’re constructing, their cross product will also be perpendicular to that plane.