Proof of angle between vectors (using dot and cross product)

**Idlewild** · January 5th 2011, 08:18 PM

Prove that if $\theta$ is the angle between u and v and u . v =/= 0, then tan $\theta$ = ||u x v||/(u . v)

where . is dot product

I am pretty familiar with the properties of both dot and cross product but that tan throws me for a loop, I just can't seem to find a way to do this

**Drexel28** · January 5th 2011, 08:22 PM

Originally Posted by Idlewild

Prove that if [Theta] is the angle between u and v and u . v =/= 0, then tan [Theta] = ||u x v||/(u . v)

where . is dot product

I am pretty familiar with the properties of both dot and cross product but that tan throws me for a loop, I just can't seem to find a way to do this

Recall that $\displaystyle \sin(\theta)=\frac{\|a\times b\|}{\|a\|\|b\|}$ and $\displaystyle \cos(\theta)=\frac{a\cdot b}{\|a\|\|b\|}$ .

**Idlewild** · January 5th 2011, 08:32 PM

Thanks for the speedy reply!

I'm not sure if this is too basic, or only physics related or something but I guess that means to use SOH CAH TOA, where O = || a x b ||, and A = a . b

Therefore tan $\theta$ = || a x b || / a . b
and ah! that turns out to be the given answer!

now to prove it... I see that u . v =/= 0, which means that u and v are not orthogonal.. That's all I really can come to right now, I haven't had much luck with the geometric interpretation

Edit: Or am I complicating this, and simply showing the explanation of sin and cos to tan would be sufficient?

**pickslides** · January 5th 2011, 09:07 PM

You mean saying $\displaystyle\tan\theta = \frac{\sin \theta}{\cos \theta} =\frac{\frac{\|a\times b\|}{\|a\|\|b\|}}{\frac{a\cdot b}{\|a\|\|b\|}} = \frac{\|a\times b\|}{a\cdot b}$ ?

**Idlewild** · January 5th 2011, 09:09 PM

Yes, I was going to show the SOH CAH TOA thing, but that works.
Do you think that would be acceptable?

**pickslides** · January 5th 2011, 09:12 PM

Maybe it would be acceptable, depends on your lecturer, you may need to prove

$\displaystyle \sin(\theta)=\frac{\|a\times b\|}{\|a\|\|b\|}$ and $\displaystyle \cos(\theta)=\frac{a\cdot b}{\|a\|\|b\|}$

as well for the answer to be accepted, this brings you back to square one.

**Drexel28** · January 5th 2011, 09:16 PM

Originally Posted by pickslides

Maybe it would be acceptable, depends on your lecturer, you may need to prove

$\displaystyle \sin(\theta)=\frac{\|a\times b\|}{\|a\|\|b\|}$ and $\displaystyle \cos(\theta)=\frac{a\cdot b}{\|a\|\|b\|}$

as well for the answer to be accepted, this brings you back to square one.

You don't really "prove" these though. By definition one takes $\displaystyle \cos(\theta)=\frac{a\cdot b}{\|a\|\|b\|}$ and the other one is gotten from the identity $\sin^2(\theta)+\cos^2(\theta)=1$ .

**Idlewild** · January 5th 2011, 09:49 PM

It is the first part of the first question on this assignment and I do also think that we are given the definition for the cos, so I will just show by identities

Thanks everyone

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