MHB Calculating the Angle Formed By Vectors

[veronica1999]

Show that there are at least two ways to calculate the angle formed by the vectors [cos 19, sin 19] and [cos 54, sin 54].

1) I can draw a unit circle and easily see that the angle is 35.

2) Change the values to decimals and use the law of cosines.
(Tried this but the calculation was a bit messy)

or use cos(theta) = uv/ lullvl

3) Use dot product and get the equation for
cos(54-19) = cos54cos19 + sin54sin19

Did my answers get the point of the problem?

 

[Reckoner]

Did my answers get the point of the problem?

I think so. The two simplest ways that I can think of would be:

1. Recognize that \(\langle\cos\theta, \sin\theta\rangle\) is a unit vector with direction \(\theta\) relative to the positive \(x\)-axis. So \(\langle\cos19^\circ, \sin19^\circ\rangle\) makes an angle of \(19^\circ\) with the \(x\)-axis, and \(\langle\cos54^\circ, \sin54^\circ\rangle\) makes an angle of \(54^\circ\). Taking the difference between the two angles, we get \(35^\circ\) as the angle between the two vectors.

2. Using the formula for the angle \(\theta\) between two vectors \(\mathbf u\) and \(\mathbf v\) gives

\[\cos\theta = \frac{\mathbf u\cdot\mathbf v}{\|\mathbf u\|\|\mathbf v\|}\]

\[\Rightarrow\cos\theta = \cos19^\circ\cos54^\circ + \sin19^\circ\sin54^\circ\]

\[\Rightarrow\cos\theta = \cos35^\circ\]

So \(\theta = 35^\circ\).

 

[Ackbach]

Another method would be to increase the dimension by one, and use the cross product. In three dimensions, the cross product and the dot product each give you a distinct way to compute the angle between two vectors.