I’m finally getting to the end of the material on vectors in my first year course: 6 planned lectures turned into 8.5. My original plan was to make systematic use of the following lemma: if is a unit length vector then expresses as a linear combination of a vector parallel to and a vector perpendicular to . In particular, the `length of in the direction of ‘ is .
Typed out here this still seems to make good sense, and I find it very helpful to think primarily in terms of the projection map . But I think the lemma was found to be unmotivated and so people switched off for the proof; it then never seemed right to use it in lectures. Perhaps the problem was introducing it too soon after the definition of the dot product. Anyway, instead I ended up running the argument to prove the lemma each time. My conclusion, for the moment, is that this wasn’t such a bad thing.
Today (a double billing, I started writing this between lectures) the lemma was used to find the distance between two skew-lines and the volume of a parallelepiped. For the latter, I made a fresh attempt at isolating the idea, in a new, more specialized lemma, and I think it went down much better.
Let be a right-angled triangle with hypotenuse . Let be a vector of unit length parallel to . The length of is .
It’s easy to complicate the proof by juggling with scaling factors. Of course any ‘extra’ lines are utter trivialities, but when every step halves the chance of the students getting anywhere …
We have . Here is parallel to and is perpendicular. Hence
Let , so (1) and (2)
. By (1), then (2), then the displayed equation, we get
Applications of the new-style lemma.
Distance of point from plane through origin with unit normal
The closest point makes a right-angled triangle exactly as in the lemma.
Volume of parallelepiped formed by , ,
The volume of the base is , and the height is where and is the closest point in the plane through the origin with normal .
Interpretation of in
No triangle is necessary, but the idea is close to the proof: if is the closest point in to the origin then, since is parallel to , we have .
Problems that do not fit
Distance of point from a line through the origin
If is the closest point then we now know the direction of (since is in the direction of the line) rather than . So it’s just sufficiently different from the lemma to be potentially confusing.
Distance between two non-parallel lines
If the lines go through and and have directions and then the vector between the two closest points is in direction , and the shortest distance is
The final cancellation is nice to see, so I wouldn’t even want to use the lemma here.