Equation for a plane perpendicular to a line through two given points 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.
 A: 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
$$x + 2y = 2.$$
A: You don't need the equation of the line, only its direction vector, for this vector is the normal vector of the plane.
$$\vec n=(2,1,0)-(1,-1,0)=(1,2,0)$$
Now, the equation of the plane is
$$x+2y=C$$
To find $C$ you only have to subst the point the plane passes through.
$$0+2\cdot1=C\implies C=2$$
A: A direction vector of the line is $$\begin{pmatrix}2-1\\1--1\\0-0\end{pmatrix}=\begin{pmatrix}1\\2\\0\end{pmatrix}$$
If the plane is perpendicular to the line, the normal vector of the plane is equal to the direction vector of the line (convince yourself of this).
A plane equation with normal vector $(a,b,c)$ and passing through the point $(x_0,y_0,z_0)$ is $a(x-x_0)+b(y-y_0)+c(z-z_0)=0$.
$$(x-0)+2(y-1)=0 \Longleftrightarrow x+2y=2$$
