I have my biggest maths test tomorrow, and I have this one problem. I understand how to find the shortest distance between a point and a plane using vector projections.
however, I do not comprehend as to how we get from vector projections formula (magnitude of the projection) to the formula for distance between the point and the plane. And also, how is that related to distance between two parallel planes, if it is?
this is the formula: http://s3.amazonaws.com/answer-board-image/cramster-equation-20113281541336343692369360972354302.gif
Thank you so much in advance.
This, I presume you already know:
Note that a plane is defined through 3 non-colinear points, any other definition is a twist of this one (though you could argue the other way, but this defition, to me, is the most "raw")
First we have a vector $k$ that takes us from the origin to a point on the plane. Then there are 2 other vectors that take us from that point to the 2 other points. Call them $q$ and $m$.
Now, we can define our plane parametrically using the equation $t = k + Oq + Um$, where $t$ is an arbitrary point on the plane and $O$ and $t$ are independent parameters.
If you don't know why this is true, I really suggest you look it up (plane parameterization).
Now, we know there will be a certain point on the plane given by vector $t$, which an arbitrary point $l$ (outside the plane) will be the closest to. Now there will be a vector $l-t$ connecting these 2 points.
The special property of this connective vector is that it is normal to the plane, always.
Because if it wasn't, then you could construct a triangle with the hypotenuse being the initial "shortest" vector, but then there would be a demonstrably closer point on the plane to point $l$ (given by the adjacent side).
Alright, now we have enough information to solve such problems. So for a given $l-t$, there will be 2 known vectors (the ones multiplied with the parameters in the parametric equation) what will be perpendicular.
Note: the dot product will be 0 in both cases. Also now, $l-t$ will be a vector of 2 variables (the parameters).
So now you have 2 equations and 2 variables, solve the system for the 2. From that, find $t$. Now apply the distance formula to get the shortest distance.
To your actual question. The formula you have shown is a very useful "compactification" of this whole process. If you where to express the parameters $O$ and $U$ in the final equation in terms of the values from which they have been derived, you would end up with one monstrous algebraic expression for the distance between a plane and a point.
Now, I suspect what the have done in the link above is the greatly, and I guess skillfully, simplified the formula.
As for the 2 parallel planes bit. Note, we can define a new plane $t = k + Oq + Um - t + l$ which goes to $l = k + Oq + Um$. This new plane will be parallel (non intersecting) to the original plane but it will be translated by vector $l - t$. Hence, given a point in the first plane, the closest point in the newly defined plane will be given by the vector $l - t$, always.
Also (this is the direct answer to what you asked), the shortest distance between a point in one plane and the other plane will always be a constant (in fact, it is the very distance between the 2 planes), no matter what point you were to choose for the one plane.
If you draw all of this out, you will see it's nothing really bad!
I hope this helped, even if only a bit. Ask if you are unsure about anything!
Good luck on your test :)