Vector product and angles

[jegues]

Homework Statement

Two vectors lie their tails at the same point. When the angle between them is increased by 20 degrees the magnitude of their vector product doubles. The original angle between them was about _____ ?

Homework Equations

\overline{a}\bullet\overline{b} = |\overline{a}| |\overline{b}| cos\theta

The Attempt at a Solution

Let the original angle be c and let the final angle be d, defined as follows:

d = 20 + c;

a * b = |a||b| cos(c)

a * b = 2[ |a| |b| ] cos(d)

cos(d) = cos(20 + c) = cos(20)cos(c) - sin(20)sin(c)

I get stuck about here, I can't figure it out...

 

[MaxL]

I think the author of the problem wants you to do it numerically. That's why it says 'about.'

 

[gabbagabbahey]

Let the original angle be c and let the final angle be d, defined as follows:

d = 20 + c;

a * b = |a||b| cos(c)

a * b = 2[ |a| |b| ] cos(d)

cos(d) = cos(20 + c) = cos(20)cos(c) - sin(20)sin(c)

I get stuck about here, I can't figure it out...

Okay, so now you basically have

\cos x=2\left(\cos20^\circ\cos x-\sin20^\circ\sin x\right)[/itex] <br /> <br /> (using x instead of &#039;c&#039;)<br /> <br /> Can you think of a relation between \sin x and \cos x that would allow you to express this equation as a polynomial in the variable u\equiv\cos x?

 

[MaxL]

Also, doesn't 'vector product' refer to the cross product?

 

[jegues]

Can you think of a relation between and that would allow you to express this equation as a polynomial in the variable ?

The only thing I can think of is if u = cosx, then -du = sinx. How can I end solving my u with that du sticking around... You might just have to reword that last line for me ;)

EDIT: This is ruining me! The only only relevant relationship would be sin^2 + cos^2 = 1 identity. If I had any hair on my head I'd be tearing it out by now.

 

[ideasrule]

Don't get du involved because this is not an integration problem, just an algebra problem.

sin^2 x + cos^2 x= 1 is almost certainly what MaxL was thinking of! If u=cosx, u^2=cos^2x. What would sin^2 x be?

 

[ideasrule]

BTW, I agree with MaxL: the question is intended to be solved numerically, i.e. using a graphing calculator or by guessing different values of theta.

 

[jegues]

sin^2 x + cos^2 x= 1 is almost certainly what MaxL was thinking of! If u=cosx, u^2=cos^2x. What would sin^2 x be?

sin^2x = 1-u^2; I don't have a sin^2 x though, can I get it somehow from sin(20)sinx, I don't know how to turn this into sin^2(BLAH) ?

 

[gabbagabbahey]

sin^2x = 1-u^2; I don't have a sin^2 x though, can I get it somehow from sin(20)sinx, I don't know how to turn this into sin^2(BLAH) ?

Isolate \sin x on one side of the equation, and then square both sides...

Also, as MaxL pointed out, "vector product" is usually interpreted as the cross product, not the dot product...so if that's what the question asks for...

 

[gabbagabbahey]

BTW, I agree with MaxL: the question is intended to be solved numerically, i.e. using a graphing calculator or by guessing different values of theta.

I don't necessarily agree with this...an exact answer is possible by means of the quadratic formula..so unless this is part of a unit on numerical analysis, I think that by using the word "about" the questioner is just asking for a numerical value (the exact answer is arccos(some garbage) ), not that the problem be solved by numerical methods.

 

[MaxL]

I stand corrected! There is absolutely a way to get the answer in the form of Arctan(stuff you know). Just remember Sin/Cos=Tan. I did it in about five lines of algebra, and didn't have to use the quadratic formula.

Hang in there, Jegues!

Also...vector products are cross products. So the dot product relation you put in your original post won't help you.

 

[ideasrule]

(slaps self in face) Of course! Can't believe I missed that.

 

[jegues]

So your solving this line for sinx,

\cos x=2\left(\cos20^\circ\cos x-\sin20^\circ\sin x\right)

I got Tan^-1(BLAH) = -68.7, again clearly not the answer... D:

 

[gabbagabbahey]

So your solving this line for sinx,



I got Tan^-1(BLAH) = -68.7, again clearly not the answer... D:

If you show us your work, we should be able to see where you are going wrong...

 

[MaxL]

Jegues-you're still using dot product relations. The problem is talking about cross products.
|(a x b)| = |a||b|Sin(theta)

You're close!