This is the question I am trying to solve: Let $V = (0,3,2)$ and $A = (-1,0,1).$ Let $P$ be the point on the line passing through $A$ with direction vector $(1,1,1)$ that is closest to $V.$ Find the distance between $V$ and the line.
This is the diagram:
I know that P is (2/3, 5/3, 8/3) and that $\overrightarrow{AP}$ is (5/3, 5/3, 5/3). However, I am still having trouble answering this question because I don't even know how to start. Any advice/help?
A standard way to find the distance without having to calculate the closest point $P$ is as follows:
Project $\vec{AV} = (1,3,1)$ onto the direction vector $d=(1,1,1)$:
$$\left(\vec{AV}\cdot \frac d{|d|}\right)\frac d{|d|}$$
Then, you have
$$|\vec{PV}|^2 = |\vec{AV}|^2 - \left|\left(\vec{AV}\cdot \frac d{|d|}\right)\frac d{|d|}\right|^2 = |\vec{AV}|^2 -\frac{|\vec{AV}\cdot d|^2}{|d|^2}$$
$$= 11 - \frac{25}{3}\Rightarrow |\vec{PV}|=\sqrt{\frac 83}$$