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Ok, so we know that if we have an arbitrary point, $p$, and a normal perpendicular to an arbitrary plane, $n$, the distance from the point to the plane can be computed as follows:

$$distance = p \cdot n - d $$

where the operator '$\cdot$' represents the dot product between two vectors, and $d$ is the distance of the plane itself from the origin.

What I'm not sure about, though, is the visualization of this in my head.

We know that, given two vectors $a$ and $b$:

$$a \cdot b = \cos\theta |a||b| $$

where $\theta$ is the angle between $a$ and $b$; while $|a|, |b|$ are the magnitudes of $a$ and $b$, respectively.

So, I get that. Take the dot product of two vectors, and the cosine itself is simply scaled by the product of the resulting magnitudes (unless they both are normalized, in which case just the cosine is the result itself).

However, in the context of computing the distance from a point to a plane, is it more appropriate to visualize the point as a vector whose tail stems from an arbitrary world-origin in 3D space, or a vector whose tail stems from an arbitrary point within the plane itself?

Of course, the point itself is the head of the vector to visualize here.

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2 Answers2

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Refer to the following diagram:

enter image description here

It seems more immediate to regard the point $P$ as being represented by the vector $p=\overrightarrow{OP}$ since it uses the coordinates of $P$ directly. Then if $n$ is a normed normal vector to the plane $\alpha$ containing the point $S$ it is also the normal vector of a parallel plane $\beta$ containing the origin.

This makes $|p\cdot n|$ equal to the distance $\operatorname{dist}(P,\beta)=\operatorname{dist}(P,\alpha)+d$ where $d$ is the distance between $\alpha$ and $\beta$. In the above example $p\cdot n\approx 13.28$ and $d\approx 8.66$. BTW. $\overrightarrow{OS}=(5,-5,5)$ is parallel to the normal vector $n\approx(0.58,-0.58,0.58)$ in this example. Here $0.58\approx 1/\sqrt3$ which pops out when normalizing the vector $(5,-5,5)$.

Of course you could have drawn a vector $p'=\overrightarrow{SP}$ and calculated $\operatorname{dist}(P,\alpha)=|p'\cdot n|$ directly.

I hope this answers your question - otherwise comment to ask me to elaborate further.

String
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I guess I think of it as tail at the point, and the head at another arbitrary point in the plane.

Then, the projection of that vector, onto the normal line ( from same point to plane) is the distance.

a \dot b

but is really:

a hat dot b hat

so think about the difference that the order makes on the projection that you describe.

That's my 2 cents, worth every penny, haven't been in school for a while..

wbg
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