# Cross Products, Dot Products, Gradients, and Divergence

• Oct 12th 2012, 07:39 PM
Brendan
Cross Products, Dot Products, Gradients, and Divergence
Ok a quick summary. I am in a class (3 weeks in) and my professor is horrible at being a teacher. So far every class has consisted of non stop proofs on the chalk board with bad handwriting, the teacher has yet to write a number on the board, we have done no problems together as a class, and he never asks if we have any questions. The homework is printed out and is not from the book, and there is no online syllabus or information.

I just took an Exam today on material that we have never turned in homework for... and I was lost on half the problems. Would really love to have some help if anyone knows this stuff. I have spent hours studying the book, and although I am learning a lot from there, it doesn't seem to relate directly to whatever the hell is happening on the board.

Here is my current homework, and the 2 problems I wrote at the bottom are from the exam, the ones I didn't know how to begin. Thanks!
• Oct 12th 2012, 09:09 PM
chiro
Re: Cross Products, Dot Products, Gradients, and Divergence
Hey Brendan.

Are you stuck with all of the problems? I'll start off with the ones you wrote.

Prove |AXB|^2 + (A.B)^2 = |A|^2|B|^2

Now |AXB|^2 = sin^2(theta)*|A|^2*|B|^2 since we have AXB = n*sin(theta)*|A|*|B|. If we take the magnitude of this vector and square it, we get

|AXB| = |n*sin(theta)*|A|*|B|) = |n|*|sin(theta)*|A|*|B|| but n is a unit vector which means |n| = 1 so we get

|AXB| = |sin(theta)|*|A|*|B| and squaring gives us

|AXB|^2 = sin^2(theta)*|A|^2*|B|^2.

Now for the dot product we know that <A,B> = |A|*|B|*cos(theta) so squaring this gives

(<A,B>)^2 = |A|^2|B|^2*cos^2(theta).

Adding the two gives us

|A|^2|B|^2sin^2(theta) + |A|^2|B|^2cos^2(theta) but sin^2(theta) + cos^2(theta) = 1 which means the total is

|A|^2|B|^2(1) = |A|^2|B|^2.

If you weren't introduced to the sin(theta) form for AXB this would have been very confusing.

For del(F), you need to specify what F is explicitly: you have given r but r by itself doesn't mean anything. Please describe what that the r vectors mean (it gives a definition for phi in Q3 but not for r and I don't know why).
• Oct 12th 2012, 09:18 PM
Brendan
Re: Cross Products, Dot Products, Gradients, and Divergence
Thanks! I will need to rewrite that on paper, and practice it to make sure I can do it no problem. With the del(F) problem on the bottom, I am assuming it is related to the question on problem 2 near the top, which also lists an answer. I am assuming the answer to the Exam question I wrote at the bottom has a similar form of answer?
• Oct 12th 2012, 09:23 PM
chiro
Re: Cross Products, Dot Products, Gradients, and Divergence
It looks like r is just the length of the vector and r_hat is the direction vector which is really strange notation.
• Oct 12th 2012, 09:26 PM
Brendan
Re: Cross Products, Dot Products, Gradients, and Divergence
Yeah. It has been frustrating because I can't find a comparison problem in the book, and I usually can't keep pace with this guys lectures so I don't take notes of everything. He hands out an answer key after everything is graded, but so far I can't seem to get on top of this. I appreciate ya looking at it though.
• Oct 12th 2012, 09:32 PM
chiro
Re: Cross Products, Dot Products, Gradients, and Divergence
I can go over the other problems with you if you want, just show us what you are thinking and any approach you tried to take.
• Oct 12th 2012, 11:12 PM
Brendan
Re: Cross Products, Dot Products, Gradients, and Divergence
Ok, can you show me number 5?
• Oct 12th 2012, 11:45 PM
chiro
Re: Cross Products, Dot Products, Gradients, and Divergence
For number 5 we have the authors convention where x = r*r-hat where r^2 = (x^2 + y^2 + z^2) and r-hat seems to be a basis for <x,y,z> (so the vector represents a linear combination of ax + by + cz where x,y,z is like i,j,k).

First we need to find r^3*r-hat. We know that r = SQRT(x^2 + y^2 + z^2) and r_hat is just a basis.

For <d/dx,d/dy,d/dz> . <(x^2 + y^2 + z^2)^(3/2) * <x,y,z>). we use the product rule:

d/dx ((x^2 + y^2 + z^2)^(3/2)*x) = (x^2 + y^2 + z^2)^(3/2) + 2*(3/2)*x*(x^2 + y^2 + z^2)^(1/2)*x = r^3 + 3x^2*r
d/dy ((x^2 + y^2 + z^2)^(3/2)*y) = (x^2 + y^2 + z^2)^(3/2) + 2*(3/2)*y*(x^2 + y^2 + z^2)^(1/2)*y = r^3 + 3y^2*r
d/dz ((x^2 + y^2 + z^2)^(3/2)*z) = (x^2 + y^2 + z^2)^(3/2) + 2*(3/2)*z*(x^2 + y^2 + z^2)^(1/2)*z = r^3 + 3z^2*r

So we add these up (since it is a dot product) and we get

(r^3 + 3x^2*r) + (r^3 + 3y^2*r) + (r^3 + 3z^2*r) = 3r^3 + 3r*(x^2 + y^2 + z^2)
= 3r^3 + 3*r*r^2
= 3r^3 + 3r^3
= 6r^3

He is using weird notation (but still one that makes a bit of sense) in that r_hat = <x,y,z> and r^2 = x^2 + y^2 + z^2.
• Oct 13th 2012, 01:55 AM
Brendan
Re: Cross Products, Dot Products, Gradients, and Divergence
I just realized that r was radius for a sphere. Just never done one of these problems ever before. Thank you so much for solving this for me so quickly. I really needed to see the breakdown. You do very well with notation. Thanks! I am going to try to see what I can solve now.