Problem on finding a normal vector of a plane
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Find a normal vector to the plane $2x + y - 3z = 1$.
Solution
Recall the relationship between a plane and its normal vector:
A plane is specified by a point and a normal vector
A plane with normal vector $\bfn$ is given by $\bfn\cdot \bfx = b$ for some $b$.
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
Direct computation of dot product
$(n_1, n_2, n_3)\cdot (x, y, z) = n_1 x + n_2 y + n_3 z$
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 $\bfn = (2, 1, -3)$.
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