Vectors question

[Dreadnought]

|b|^2 = -9b.a/27
and
b.a/|b|^2 = -9/27

are not equal (you've flipped one side).

[Panphobia]

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

[Dreadnought]

27*(B.B) = (-9)*(B.A)
( 27 / -9 )*(B.B) = (B.A)
(-3)*(B.B) = (B.A)
-3 = (B.A)/(B.B)

[Panphobia]

then you would expand right side to be 3|b|cosQ/|b| and cos-1(-1) = 180, which is impossible???

[Dreadnought]

I've been wondering that myself for over an hour now... (rather that finishing my own assignment )

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).

[Panphobia]

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

[Dreadnought]

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

[Brightguy]

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.

[Dreadnought]

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.

[Panphobia]

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?

[Panphobia]

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?

[Dreadnought]

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.

[Panphobia]

ohhhhhhhhhhhhh so everytime you figure out work you need to reduce the force vector to a unit vector?

[DemonWasp]

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.

[Panphobia]

yea ok i understand that, but when wouldn't you reduce the direction to a unit vector?