Cosine question. Scalar product.

[LagrangeEuler]

Homework Statement

Find angle between vectors if
$$\cos\alpha=-\frac{\sqrt{3}}{2}$$[/B]

Homework Equations

The Attempt at a Solution

Because cosine is negative I think that $$\alpha=\frac{5\pi}{6}$$. But also it could be angle $$\alpha=\frac{7\pi}{6}$$. Right? When I search angle between vectors I do not need to write $$+2k\pi$$ where $$k$$ is integer. Right? Thanks for the answer.[/B]

 

[jedishrfu]

I'd say you're right. The dot product of two unit vectors giving a negative cosine just means the angle between them is greater than 90 degrees and if you look at a diagram of the two possible angles you'll see they are symmetrical about a line thru one of the vectors.

 

[LagrangeEuler]

Maybe only is important to look arccos as function? So answer is only $$\alpha=\frac{5\pi}{6}$$?
So if I look at calculator is $$\alpha=arccos(...)$$ is this $$\alpha$$ angle from $$[0,\pi]$$ or from $$[-\pi,\pi]$$.

 

[Mark44]

Maybe only is important to look arccos as function? So answer is only $$\alpha=\frac{5\pi}{6}$$?

Looks fine to me. If you have two rays that emanate from the same point, two angles are determined- a smaller one and a larger one (I'm assuming here that the two rays don't point in exactly opposite directions). For problems asking about the angle between the two rays, they're usually interested in the smaller of the two angles.