## Problem on finding a normal vector of a plane

Find a normal vector to the plane $2x + y - 3z = 1$.
• ## Solution

Recall the relationship between a plane and its normal vector:
To find $\bfn$, we should write the given line in this form.
Let $\bfn = (n_1, n_2, n_3)$ and $\bfx = (x, y, z)$.
Recall that
Thus, the plane $\bfn \cdot \bfx = b$ is the same as $n_1 x + n_2 y +n_3 z = b$.
We identify that $n_1 = 2$, $n_2=1$, $n_3 = -3$ and $b = 1$.
That is, the plane can be written in the form $$(2, 1, -3)\cdot(x,y, z) = 1.$$
We conclude that a normal vector to the plane $2x+y-3z=1$ is $\bfn = (2, 1, -3)$.