# 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?

• Looks like it is asking for the distance from the plane to the cat's head, the distance from the plane to the cat's tail, and then you compare those two values. Are you familiar with normal (i.e. perpendicular) vectors to a plane? Jul 15, 2020 at 2:48
• Yes, I've encountered normal vectors once or twice, but I am not the best with solving problems including them. Jul 15, 2020 at 2:51

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 $$(3, 2,1)$$.

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.

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.

• Oh I see, I just plug in the values of both coordinates, find the difference, and that is my answer. Thank you! I will try this and hopefully get it right this time. Jul 15, 2020 at 2:55
• I just solved it and got (13sqr(14))/14, which is correct. Thank you! Jul 15, 2020 at 3:04