Find an equation of the plane containing the point $(1, 1, -1)$ and perpendicular to the line through the points $(2, 0, 1)$ and $(-1, 1, 0)$
This is what I have:
I first find the vector between the points: $\vec{n}^{\ } = (2,0,1) – (-1, 1, 0) = (3, -1, 1)$
Next I find the formula for the plane that is perpendicular to the vector by taking the dot product of the vector and the given point + another point of the plane
$(3, -1, 1) \cdot (1 + x, 1 + y, z -1 )\\ 3 + 3x -1 – y -1 -z\\ 3x – y – z +1 = 0$
I am not at all sure if I followed the correct steps
Best Answer
The point-normal form of a plane is $\vec n \cdot(p - p_0) = 0$. If you think about the meaning of this, you will find that for any point $p$ on the plane, if you form a vector from that point and a point known to be on the plane, $p_0$, that vector will be orthogonal to your normal vector $\vec n$.