In class we tried to find the distance from to the plane of equation .
There are several ways of doing this. For example:
- Fix a point on the plane . Any point will do, say .
- Find a vector perpendicular to . For example, .
[Again, if a plane has equation then is perpendicular to it.
For example, is parallel to the plane , which we can rewrite as , which is the equation of the set of points perpendicular to the vector . This is to say, the plane is perpendicular to the vector . Since the plane : is parallel to , is also perpendicular to .
Another way of reaching the same conclusion is to rewrite in the form for some appropriate vector . There are many choices of and they all work; all we need is that , i.e., that is in the original plane. For example, the point belongs to the plane : , so we can rewrite the equation of as , which is equivalent to saying that But this means that the vector is perpendicular to the vector , which is an arbitrary vector in the direction of the plane.]
Let’s continue with the problem of finding the distance from to :
- Consider the projection of the vector in the direction of . Clearly, the distance from to is the length of .
- Recall that .
- Then and the distance is .
- In detail, so and , so .
Notice that (as discussed in class) the distance from a point to the line in the direction of that goes through a point is given by , while (by the above) the distance from a point to a plane containing a point and perpendicular to a vector is given by . While the expressions are similar, one involves a cross product and the other a dot product. This is because in one case we express the distance in terms of the sine of an angle , and in the other, in terms of its cosine or, what is the same, in terms of the sine of . (Drawing a diagram may help you clarify the situation.)
Have we gotten our Homework 2 assigned yet?
HW 2 is posted now under syllabus.
Thanks a bunch