Difference between distances of two points from a plane This is the question I am trying to solve: My cat is standing on the plane $3x+2y+z = 4.$ Its head is at $(0,-1,-3)$ and the tip of its tail is at $(-4,-2,-2).$ Measured from the plane, what is the difference between the height of my cat's head and my cat's tail?
I originally just thought the answer was just the distance between the two points, and I got ${3 \sqrt{2}}{},$ however, it is wrong. I don't know what else to do. Any ideas?
 A: The twist here is that the cat is not standing perpendicular to the plane and is along the vector $(4, 1,-1) $. So, you have to project the length $3\sqrt 2$ along $(4, 1,-1) $ on the normal vector $(3, 2,1) $.
A: The distance of a point $Q(x_0,y_0,z_0)$ from the plane $3x+2y+z-4=0$ is given by
$$d_Q=\frac{|3x_0+2y_0+z_0-4|}{\sqrt{3^2+2^2+1^2}}.$$
Now calculate the distance $d_h$ of cat's head and that of tail $d_t$ from the plane, then find the difference.
A: Your question was already answered bu since I, personally, never remember the formula, I want to show you how to derive it. You have to be familliar with plane and line ecuations (cartesian and parametric), dot product and norm.
Let your plane $\pi: Ax + By + Cz = D$ and your point $\vec{P} = (x_p,y_p,z_p)$. Another way to write the ecuation for your plane is
$$\pi: \vec{n}\cdot\vec{X} = \vec{n}\cdot\vec{Q} $$
where $\vec{n} = (A,B,C)$ is the normal vector, $\vec{X} = (x,y,z)$ and $\vec{Q}$ is a point that is on the plane (this isn't important for what's next, remember that $\vec{n}\cdot\vec{Q} = D$).
You should agree with me that the point in the plane closer to $\vec{P}$ (will call it $\vec{R}$) passes throught the line defined by the point $\vec{P}$ and the vector $\vec{n}$ (line perpendicular to the plane passing throught $\vec{P}$). If you have trouble seeing this, draw it in 2D: it's analogous in more dimentions. So we can write $\vec{R}$  as $$ \vec{R} = \lambda\,\vec{n} + \vec{P}$$
with $\lambda\in\mathbb{R}$.
But this point is also contained in the plane. Therefore, it satisfies our plane ecuation.
$$\begin{align}
\vec{n}\cdot\vec{R} &= D \\
\vec{n}\cdot(\lambda\,\vec{n} + \vec{P}) &= D  \\
\lambda(\vec{n}\cdot\vec{n}) &= D -(\vec{n}\cdot\vec{P}) \\
\lambda &= \frac{D -(\vec{n}\cdot\vec{P})}{\vec{n}\cdot\vec{n}} \\
\end{align}$$
We have found our point $\vec{R} = \frac{D -(\vec{n}\cdot\vec{P})}{\vec{n}\cdot\vec{n}}\,\vec{n} + \vec{P}$
Now, you only have to apply the norm to the difference of $\vec{P}$ and $\vec{R}$.
$$\begin{align}
\|\vec{R}-\vec{P}\| &= \|\frac{D -(\vec{n}\cdot\vec{P})}{\vec{n}\cdot\vec{n}}\,\vec{n} + \vec{P} - \vec{P}\| \\
&= \|\frac{D -(\vec{n}\cdot\vec{P})}{\vec{n}\cdot\vec{n}}\,\vec{n}\| \\
&= |D -(\vec{n}\cdot\vec{P})|\,\frac{\|\vec{n}\|}{\|\vec{n}\|^{2}} \\
&= \frac{|D -(\vec{n}\cdot\vec{P})|}{\|\vec{n}\|} \\
\end{align}$$
Which you can re-write as
$$d(\vec{P},\pi) = \frac{|Ax_p + By_p + Cz_p - D|}{\sqrt{A^2+B^2+C^2}}$$
I don't know if I there is a shorter way. If there is any, please comment.
