Question

Hi guys, Matt from Australia here. Brushing up on course material before beginning Bachelor of Science in Physics next year through an Australian university. With this subject, I found that they have chopped up the old lectures from 1999 a bit. I have instead downloaded all 20-something lectures and am watching them in order. It seems to be working fine. I'd like to get this subject knocked over in the next 6 weeks or so, I'd be interested to hear from anyone on the same timeline as me. Two questions at the moment...

Answer 1

1) in the challenge problems for vectors (topic 3), question 1.4 shows the answer to be 63 deg... I got 56 deg. Anyone have this issue also?

Answer 2

2) same topic (vectors, topic 3) challenge problems, problem 2.1. The notes show that the denominator of the function is magnitude of vector A * magnitude of vector B, yet when he calculates it out, he uses root3 * 2 (i.e. root3 + root 3, instead of root 3 * root 3 = 3). Hope this makes sense.

Answer 3

3) and another one... maybe I'm just not getting it... see attachment and red circled areas. Cross product formula for Bx should be -2?

Answer 4

For your question 1) the challenge problems for vectors (topic 3), question 1.4 .
I also get answer 56 degree.
I think the solution in the pdf has mistake.

Answer 5

For your question 2 about vectors( topic 3) challenge problems, problem 2.1,
I think the solution in the pdf has mistake again.
The angle should equal to : arc cos ( -1 / ( 3^0.5 X 3^ 0.5) )
= arc cos ( -1/3 )
= 109 degree

reference:
http://en.wikipedia.org/wiki/Tetrahedral_molecular_geometry

Answer 6

I am programmer and computer science usually use ^ to represent power.
e.g. square root of 3 = 3 power 0.5 = 3^0.5

Answer 7

For your question 3 about vectors( topic 3) challenge problems, problem 3.1,
you are ture that the suggested answer has mistake.
Bx should be -2.

A x B
= ((1)(3)-(-1)(-1)) i + ((-1)(-2)-(1)(3)) j+ ((1)(-1)-(1)(-2))k
= 2i -j +k ............ (ans1)

Check the intermediate answer ans1:
Let C = A x B,
then
i) dot product of A and C is 0 ( becasue they are prependicular )
ii) dot product of B and C is also 0 ( becasue they are prependicular )

product of C and A
= (Ax )(Cx) + (Ay)(Cy) + (Az)(Cz)
= (1)(2) + (1)(-1) + (-1)(1)
= 0

product of C and B
= (Bx)(Cx) + (By)(Cy) + (Bz)(Cz)
= (-2)(2) + (-1)(-1) + (3)(1)
= 0


If you use the pdf's answer C = A x B = 2 i- 5 j- 3 k
to calcualte product of B and C, you will find that it is not equal to 0 !
(that means they are not prependicular):
= (Bx)(Cx) + (By)(Cy) + (Bz)(Cz)
= (-2)(2) + (-1)(-5) + (3)(-3)
= -8 !

Answer 8

Great, thanks Ricky, good to have my thoughts confirmed :)