Computer Science Canada

Vectors question
Author:  Panphobia [ Tue Mar 05, 2013 10:38 pm ]
Post subject:  Vectors question
So I have a vectors test tomorrow, and I understand all the questions for review but could not get the final answer for one thinking question here is the question,
code:
A and B are vectors
Two vectors A + 3B and 4A - 3B make an angle of 90 degrees when tail-to-tail Find the angle between A and B if |A| = 3|B|. (Hint: u dot u = |u|^2)
I started out by taking the dot product of A+3B and 4A-3B which gave me (period is dot in dot product) 4A . A + 4A . 3B - 3B . A - 3B . 3B = 0 , I am pretty sure you can simplify this more until you get to |a||b|cos(theta) but I am not sure how, can anyone steer me in the right direction?
Author:  michaelp [ Tue Mar 05, 2013 10:55 pm ]
Post subject:  RE:Vectors question
Your first step is right. Try and use the hint to simplify what you have more.
Author:  Panphobia [ Tue Mar 05, 2013 11:05 pm ]
Post subject:  RE:Vectors question
|4A|^2 + 4A . 3B - 3B . A - |9B|^2=0
lost now
Author:  Dreadnought [ Tue Mar 05, 2013 11:10 pm ]
Post subject:  Re: Vectors question
Hopefully, you know that the dot product is linear and commutative, that is
code:

For any vectors A, B and scalars s, t (real numbers) we have
(sA).(tB) = (st)(A.B)
and
A.B = B.A


As an exercise, use the definition of the dot product to prove these.
Author:  Panphobia [ Tue Mar 05, 2013 11:17 pm ]
Post subject:  RE:Vectors question
that was actually one of my projects, um could |3B| be turned to 3|B| not sure
Author:  michaelp [ Tue Mar 05, 2013 11:23 pm ]
Post subject:  RE:Vectors question
Note that |4A|^ 2 != 4|A|^2, and |9B|^2 != 9|B|^2.
Author:  Dreadnought [ Tue Mar 05, 2013 11:26 pm ]
Post subject:  Re: Vectors question
Panphobia wrote:

that was actually one of my projects, um could |3B| be turned to 3|B| not sure


Just go by definition
We define |A|^2 := (A.A)
then apply linearity to |sA|

EDIT: Also, because of how |A| is defined, it is always nicer to work with |A|^2, in this case you can do this (if you are careful, though not required).
Author:  Panphobia [ Tue Mar 05, 2013 11:33 pm ]
Post subject:  RE:Vectors question
No wonder most of the class didnt even get one mark on this question last year, psh
Author:  Panphobia [ Tue Mar 05, 2013 11:44 pm ]
Post subject:  RE:Vectors question
ok so i did this so far, am I right?
4a.a - 3b.a + 12a.b - 9b.b=0
4|a|^2 - 15b.a - 9|b|^2=0
4|3b||3b| - 15b.a - 9|b|^2 = 0
36|b|^2 - 15b.a - 9|b|^2 = 0
Author:  Dreadnought [ Tue Mar 05, 2013 11:51 pm ]
Post subject:  Re: Vectors question
Careful
- 3(B.A) + 12(A.B) = 15(A.B) ?
Author:  Panphobia [ Tue Mar 05, 2013 11:54 pm ]
Post subject:  RE:Vectors question
oh yea
Author:  Dreadnought [ Tue Mar 05, 2013 11:56 pm ]
Post subject:  Re: Vectors question
You have (-3)(A.B) + (12)(A.B) = (-3 + 12)(A.B) = ???
Author:  Panphobia [ Wed Mar 06, 2013 12:01 am ]
Post subject:  RE:Vectors question
ok i think I am completely done, point out errors please
(4a-3b).(a+3b)=0
4a.a - 3b.a + 12a.b - 9b.b=0
4|a|^2 + 9b.a - 9|b|^2=0
4|3b||3b| + 9b.a - 9|b|^2 = 0
36|b|^2 + 9b.a - 9|b|^2 = 0
27|b|^2 = -9b.a
-1/3 = b.a/b.b
-1/3 = |b||a|cosQ/|b|^2
-1/3 = |a|cosQ/|b|
-1/3 = 3|b|cosQ/|b|
-1/3 = 3cosQ
cosQ = -1/9
Q = 96.4
Author:  Dreadnought [ Wed Mar 06, 2013 12:02 am ]
Post subject:  Re: Vectors question
Almost, check this step carefully

27|b|^2 = -9b.a
-1/3 = b.a/b.b
Author:  Panphobia [ Wed Mar 06, 2013 12:10 am ]
Post subject:  Re: Vectors question
27|b|^2 = -9b.a
-1/3 = b.a/b.b

ok look either way you put it
|b|^2 = -9b.a/27
and then wouldnt b.a go b.a/|b|^2 = -9/27? or am I confused lol
Author:  Dreadnought [ Wed Mar 06, 2013 12:14 am ]
Post subject:  Re: Vectors question
|b|^2 = -9b.a/27
and
b.a/|b|^2 = -9/27

are not equal (you've flipped one side).
Author:  Panphobia [ Wed Mar 06, 2013 12:16 am ]
Post subject:  RE:Vectors question
then you explain how rearranging lets say I = V/R and R = V/I see you divide on the top like if 2 = 5*x, and x = 2/5, then 5 = 2/x no? i used the same principle
Author:  Dreadnought [ Wed Mar 06, 2013 12:21 am ]
Post subject:  Re: Vectors question
27*(B.B) = (-9)*(B.A)
( 27 / -9 )*(B.B) = (B.A)
(-3)*(B.B) = (B.A)
-3 = (B.A)/(B.B)
Author:  Panphobia [ Wed Mar 06, 2013 12:25 am ]
Post subject:  RE:Vectors question
then you would expand right side to be 3|b|cosQ/|b| and cos-1(-1) = 180, which is impossible???
Author:  Dreadnought [ Wed Mar 06, 2013 12:37 am ]
Post subject:  Re: Vectors question
I've been wondering that myself for over an hour now... (rather that finishing my own assignment Razz)

If you want my opinion, trust the math and don't worry too much about it. It's possible the only vectors that satisfy (4A - 3B).(A - 3B) = 0 and ||A|| = 3 ||B|| are A = B = 0 (the zero vector).
Author:  Panphobia [ Wed Mar 06, 2013 12:44 am ]
Post subject:  RE:Vectors question
oh alright that is a strange answer, you see this was a test question last semester and my friend did not get it right saying it was 0, sooooo it has to be something else
Author:  Dreadnought [ Wed Mar 06, 2013 1:24 am ]
Post subject:  Re: Vectors question
I think I've got a proof that A = B = 0

code:
Construct a right angle triangle with sides (4A - 3B) (A+ 3B) and H.
Clearly H is the difference of the two sides, H = (3A - 6B)

By the Pythagorean Theorem, ||H||^2 = ||4A - 3B||^2 + ||A + 3B||^2

||H||^2       = 13||A||^2 -36(A.B)
||4A - 3B||^2 = 17||A||^2 - 24(A.B)
||A + 3B||^2  = 2||A||^2 +6(A.B)
(note that I've used 9||B||^2 = ||A||^2 to simplify these)

Then the theorem gives
13||A||^2 - 36(A.B) = 19||A||^2 - 18(A.B)
32||A||^2 - 18(A.B) = 0
32||A||^2 - 6||A||^2cos(theta) = 0
( 6||A||^2 ) * ( 16/3 - cos(theta) ) = 0

So either
||A||^2 = 0
or
16/3 - cos(theta) = 0

Since |cos(theta)| <= 1  it is clear that the second case cannot hold. Hence ||A||^2 = 0.

||A||^2 = 0 implies ||A|| = 0 which implies A = 0 (the zero vector)

Since ||A|| = 3||B||, we have ||B|| = 0 and hence B = 0 ( the zero vector).


So there you have it, the sneaky bit is when we divide by (B.B) or (A.A) when computing the angle. We assume they are not zero, but if they are we can probably show that the angle between A and B is anything. Chances are you teacher didn't notice this (or maybe he/she knew and is just really evil!)

Using this trick of division by zero, it is probably possible to show the angle between A and B is in fact zero and that your friend was kinda right. haha

EDIT: fixed typo
Author:  Brightguy [ Wed Mar 06, 2013 2:15 am ]
Post subject:  Re: RE:Vectors question
Panphobia @ Wed Mar 06, 2013 12:25 am wrote:
then you would expand right side to be 3|b|cosQ/|b| and cos-1(-1) = 180, which is impossible???

Why do you say it's impossible? I worked it out and got the angle to be pi (radians). That just means the vectors are pointing in opposite directions.

Dreadnought @ Wed Mar 06, 2013 1:24 am wrote:
code:
13||A||^2 - 36(A.B) = 19||A||^2 - 18(A.B)
32||A||^2 - 18(A.B) = 0

Looks like that should be −6 instead of 32.
Author:  Dreadnought [ Wed Mar 06, 2013 2:51 pm ]
Post subject:  Re: Vectors question
Brightguy wrote:

Panphobia @ Wed Mar 06, 2013 12:25 am wrote:
then you would expand right side to be 3|b|cosQ/|b| and cos-1(-1) = 180, which is impossible???

Why do you say it's impossible? I worked it out and got the angle to be pi (radians). That just means the vectors are pointing in opposite directions.

Dreadnought @ Wed Mar 06, 2013 1:24 am wrote:
code:
13||A||^2 - 36(A.B) = 19||A||^2 - 18(A.B)
32||A||^2 - 18(A.B) = 0

Looks like that should be −6 instead of 32.

Right you are, I missed that. Seems that all I have proven is that ||A|| = 0 or cos(theta) = -1.

Ah, I see now. (A + 3B) is the zero vector, this was giving me trouble since the question says the angle between (A + 3B) and (4A - 3B) is 90 degrees. But if either vector is zero then no angle exists, hence I assumed they were non-zero.

Good, now I know why all this looked weird.
Author:  Panphobia [ Wed Mar 06, 2013 4:11 pm ]
Post subject:  Re: Vectors question
Ok these were the two thinking questions on the unit test, I think I did alright on them
code:
There are two unit Vectors a and b if |3a+2b|=sqrt(7) what is |a x b|
what i did for this is made a triangle with 3a 2b and sqrt(7), got the angle between a and b with the cosine law, did 180 - that angle so it is tail to tail, then did the cross product, and got the magnitude, this is question 2
code:
There are two vectors a = (14, -2k, 7) and b = (2q,-8,5s) if a and b are parallel, and |a| = sqrt(501), what are k, q, and s
what i did is got k by doing 14^2 + (-2k)^2+7^2 = 501, then rearranged for k, and since a and b are parallel, one is just a scalar multiple of the other, is this right?
Author:  Panphobia [ Wed Jun 19, 2013 9:17 pm ]
Post subject:  Re: Vectors question
I am doing some vector review and I came across this question, but my answer is 3 times the answer in the solutions page, and no matter what i do I can't seem to get it, well here is the question
code:
Find the work done if a force of 75N in the direction of r = (2,-1,-2) moves an object from G(3,-2,3) to H(7,-3,-8)?
Since Work = Force (dot) displacement, I got GH = (4, -1, -11) and then the force would be F = 75(2,-1,-2) so then the dot product of them would be
Work = (4)(150)+(-75)(-1)+(-11)(-150) = 2325 Joules
But the answer in the solution says 775 Joules, could you guys clear up who is right?
Author:  Dreadnought [ Wed Jun 19, 2013 9:20 pm ]
Post subject:  Re: Vectors question
You forgot that ||(2, -1, -2)|| = 3, (it's length is 3). Thus you've made your force vector 3 times too long.

EDIT: You should try to use unit vectors (length 1) for direction whenever possible.
Author:  Panphobia [ Wed Jun 19, 2013 9:37 pm ]
Post subject:  RE:Vectors question
ohhhhhhhhhhhhh so everytime you figure out work you need to reduce the force vector to a unit vector?
Author:  DemonWasp [ Wed Jun 19, 2013 10:44 pm ]
Post subject:  RE:Vectors question
Not necessarily. The reason here is that:

Quote:
a force of 75N


refers to the magnitude of the force, and:

Quote:
in the direction of r = (2,-1,-2)


refers to the DIRECTION of (2,-1,2), not that vector itself.
Author:  Panphobia [ Wed Jun 19, 2013 10:57 pm ]
Post subject:  RE:Vectors question
yea ok i understand that, but when wouldn't you reduce the direction to a unit vector?
Author:  DemonWasp [ Thu Jun 20, 2013 10:36 am ]
Post subject:  RE:Vectors question
If the force was given by something like

Quote:
the force vector is (7,8,9)


because that vector incorporates both direction and magnitude deliberately.

:


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