## A constant scalar projection defines a plane

Consider the set of 3d points whose component (scalar projection) along the vector $\bfv$ is the constant $c$. Show that these points form a plane, and find that plane.
• ## Solution

To begin, we write down an equality satisfied by this set of 3d points. Then, we will try manipulate it into the equation of a plane.
Let $\bfx$ be an arbitrary point in 3-space.
Recall that
We are given that the component of $\bfx$ along $\bfv$ is $c$. Hence $$c = \frac{\bfx \cdot \bfv}{|\bfv |} \tag{1}$$
Recall that
We can rewrite (1) as $$\bfv \cdot \bfx = |\bfv| c.$$
We identify $\bfn = \bfv$ and $b = |\bfv| c$.
Hence, the set of $\bfx$ that satisfy (1) form a plane with normal vector $\bfv$.