Question

Find the equation of the plane in R3 that contains the point A(2, 5, 5), and is perpendicular to the line
x = 6 - 3 t
y = -4 - 3 t
z = -3 t

Answer 1

Find a vector in the line and write down what it means for that vector to be perpendicular to any two points in the plane.

Answer 2

The vector parallel to the line is

\[\left(\begin{matrix}-1 \\ -1 \\ -1\end{matrix}\right)\]

(position and magnitude don't matter for orthogonality). (This is because all the coefficients of t are the same.)

Find two vectors orthogonal to this vector (i.e. dot product is zero); this describes the orientation of the plane. Filling in 0,1 and 1,0, we get:

\[\left(\begin{matrix}1 \\ 0 \\ -1\end{matrix}\right),
\left(\begin{matrix}0 \\ 1 \\ -1\end{matrix}\right)\]

This describes the orientation of the plane. To get the right position too, i.e. making it go through (2,5,5), translate it up by (2,5,5) and the definition of the plane (for arbitrary lambda values) is then:

\[
\left(\begin{matrix}2 \\ 5 \\ 5\end{matrix}\right)+
\lambda_1\cdot\left(\begin{matrix}1 \\ 0 \\ -1\end{matrix}\right)+
\lambda_2\cdot\left(\begin{matrix}0 \\ 1 \\ -1\end{matrix}\right)\]

In conventional plane equation notation:

\[x+y+z=12\]