We've now spent some time talking about projections and distances.
This is an attempt to summarize that in some way or another. Recall how
we found the vector projection of a vector **b** onto a vector **a**
(figure 1, to the right): we said that the length of the projection is
|**b**| cos(theta), and so, because

|**a**| |**b**| cos(theta) = **a . b**,

we can divide both sides by |
|**b**| cos(theta) = the length of the projection =
**a . b** / |**a**|

The actual vector projection is therefore a unit vector in the correct
direction times this length, that is,
Next consider the other (unlabeled) vector in the figure. This is the
*orthogonal projection* of **b** onto **a**, and its
length is (hopefully obviously) |**b**| sin(theta). Recalling that

|**a** x **b**| = |**a**| |**b**| sin(theta),

we can find this length by dividing both sides by |
|**b**| sin(theta) = |**a** x **b**| / |**a**|.

**Points and Lines**

Now, suppose we want to find the distance between a point and a line (top
diagram in figure 2, below). That is, we want the distance *d*
from the point *P* to the line *L*. The key thing to note
is that, given some other point *Q* on the line, the distance
*d* is just the length of the orthogonal projection of the vector
**QP** onto the vector **v** that points in
the direction of the line! That is, we notice that the
length *d* = |**QP**| sin(theta), where theta is the
angle between **QP** and **v**. So

Let's do an example. Suppose we want to know the distance between the
point *P* = (1,3,8) and the line
*x*(*t*) = -2 + *t*,
*y*(*t*) = 1 - 2*t*,
*z*(*t*) = -3 - *t*. We need some point ("*Q*")
on the line—let's take the
point (-2, 1, -3). Then a vector from this point on the line to the point
*P* is <3, 2, 11>.
This is the vector *QP* in the figure. We want the length
*d*, which is

|**QP** x **v**| = sqrt(400 + 196 + 64) = sqrt(660) = 2 sqrt(165)

and |
**Points and Planes**

Ok, how about the distance from a point to a plane? We'll do the same
type of thing here. Consider the lower diagram in figure 2.
Here we're trying to find the distance *d* between a point *P* and
the given plane. Again, finding any point on the plane, *Q*, we can form
the vector **QP**, and what we want is the length of the projection of this
vector onto the normal vector to the plane. But this is really easy,
because given a plane we know what the normal vector is. So we can
say

|**QP**| cos(theta) = **QP . n** / |**n**|,

so
An example: find the distance from the point *P* = (1,3,8) to the plane
*x* - 2*y* - *z* = 12. We need a point on the plane.
Hmm. There sure are a
lot of them to choose from. `:)` Let's pick something easy:
I'll pick *x* = 3, *y* = -3 and *z* = -3.
(Why? Just because they have to satisfy the
equation *x* - 2*y* - *z* = 12, and I was picking
numbers to try and keep *x*, *y*
and *z* moderately small.) Then our point
*Q* = (3,-3,-3). A vector from the
plane to *P* is **QP** = <-2, 6, 11>, so

= -25 / sqrt(6)

Why did we use the angle theta opposite the component of the vector giving the distance in the case of the line, and the angle adjacent for the plane? It all has to do with what we know: in the case of the line, we already know the vector that points along the line, so if we start doing dot or cross products with this vector, the angle that's involved will be the angle we used. Similarly for a plane, the vector associated with the plane that we know is the normal, so we're interested in angles from this vector to other vectors.

Last modified: Mon Aug 31 9:40:00 EDT 2015

Comments to:glarose@umich.edu

©2004 Gavin LaRose, UM Math Dept.