TITLE: Equation for a plane perpendicular to a line through two given points
QUESTION [7 upvotes]: The following type of question is quite popular with examiners at the institution where I study.
Find an equation of the plane containing the point $(0, 1, 1)$ and perpendicular to the line passing through the points $(2, 1, 0)$ and $(1, -1, 0)$
I start by calculating the the parametric equation of the line passing through $(2, 1, 0)$ and $(1, -1, 0)$, but after that I am lost.
I realize that I have to find the equation for the second line, and that the first plane will be perpendicular to it if the dot product of the vectors equal zero, but I cannot seem to put the pieces together.
Can someone please guide me in the correct direction for solving this type of problem?
Much appreciated.

REPLY [3 votes]: Since the line is perpendicular to the plane, so is any nonzero vector parallel to the line, including, the vector from $(1, -1, 0)$ to $(2, 1, 0)$, namely,
$${\bf n} := ( 2 - 1, 1 - (-1), 0 - 0 ) = ( 1, 2, 0 ).$$
Now, by definition any point $\bf x$ is in the plane if the vector ${\bf x} - {\bf x}_0$ from ${\bf x}_0 := (0, 1, 1)$ to $\bf{x} = (x, y, z)$ is orthogonal to ${\bf n}$, that is if
$${\bf n} \cdot ({\bf x} - {\bf x}_0) = 0. \qquad (\ast)$$
Note that this equation doesn't depend on the any of the specific points involved, so we've produced a completely general formula for the equation of the plane through a point ${\bf x}_0$ and with normal vector $\bf n$!
In our case, substituting in $(\ast)$ gives
$$(1, 2, 0) \cdot [(x, y, z) - (0, 1, 1)] = 0,$$
expanding gives
$$(1)(x - 0) + (2)(y - 1) + (0)(z - 1) = 0,$$
and simplifying gives
$$x + 2y - 2 = 0.$$
If you prefer standard form, of course this is
$$x + 2y = 2.$$