Equation for a plane perpendicular to a line through two given points

Question

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.

Answer 1

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 $$x + 2y = 2.$$

Answer 2

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.

$$0+2\cdot1=C\implies C=2$$

Answer 3

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 $a(x-x_0)+b(y-y_0)+c(z-z_0)=0$.

$$(x-0)+2(y-1)=0 \Longleftrightarrow x+2y=2$$