course Mth 275
Section 10.3:Most of the material is clear to me. The exception: computing the scalar projection of v onto w—I don’t understand the relevance of the magnitude of w to this value. It seems to me that only the direction of w and the magnitude and direction of v are relevant. No matter how long w is, it seems to me that only w’s direction matters.
That is so, and that's exactly why we have to divide by || w ||.
The direction of the w vector is defined by the unit vector w / || w ||.
The projection of v onto w has magnitude | v cos(theta) | = | v dot w | / || w ||.
A vector of magnitude | v dot w | / || w || in the direction of w is obtained by multiplying this magnitude by the unit vector in the direction of w, obtaining |(v dot w | / || w || * w / || w ||.
Remove the absolute value signs around v dot w and you correct for the possibility that the projection is in the direction opposite that of w.
The projection is therefore (v dot w) * w / || w || ^ 2.
I had difficulty with two of the problems, #54: “Fred and his son Sam are pulling a heavy log along flat horizontal ground by ropes attached to the front of the log. The ropes are 8 ft long. Fred holds his rope 2 ft above the log and 1 ft to the side, and Sam holds his end 1 ft above the log and 1 ft to the opposite side. If Fred exerts a force of 30 lb and Sam exerts a force of 20 lb, what is the resultant force on the log?” I’m sure I’m missing something basic, but I got nowhere with this one.
Find the sum of two vectors, the first of magnitude 30 lb in the direction of the vector whose magnitude is 8 ft and whose y and z components are respectively 1 ft and 2 ft (axes are labeled arbitrarily; with this labeling the x component is sqrt( 8^2 - 2^2 - 1^2 ) ft and the direction vector will be the corresponding unit vector). The second will be of magnitude 20 lb and the direction similarly defined, in this case by a vector of magnitude 8 ft with y and z components 1 ft and 1 ft.
The other problem I got stuck on was #66. Part a. was no problem [show that (v+w) dot (v+w) = ||v||^2 + ||w||^2 + 2(v dot w]. Part b. says to use part a to prove the triangle inequality: ||v+w|| <= ||v|| + ||w||. I was able to manipulate the equation to come up with: for the triangle inequality to be true, 2(v dot w) <= 2||v+w||, but this took me nowhere.
|| v + w || ^ 2 = (v + w) dot (v + w) = | ||v||^2 + ||w||^2 + 2 v dot w | <= ||v||^2 + ||w||^2 + |2 v dot w|, since 2 v dot w is either positive or 0 (in which case equality holds) or negative (in which case the < holds).
( || v || + || w || )^2 = || v || ^ 2 + || w || ^ 2 + 2 || v || || w ||.
|2 v dot w| = 2 || v || || w || cos(theta) <= 2 || v || || w ||.
Let me know if you have trouble putting this together to get the final result.