Question

Find the angle between the given vectors to the nearest tenth of a degree. u = <2, -4>, v = <3, -8>

Answer 1

@satellite73

Answer 2

You can try using the dot product, which has two definitions, assuming \(\vec{u}=\langle u_1, u_2\rangle \) and \(\vec{v} = \langle v_1, v_2 \rangle\):
\[\large (1) \, \, \vec{u} \cdot \vec{v} =u_1v_1+u_2v_2\]
and
\[\large (2) \, \,\vec{u} \cdot \vec{v} = \|\vec{u}\| \, \|\vec{v}\|\cos \theta =\sqrt{u_1^2+u_2^2}\sqrt{v_1^2+v_2^2}\, \cos \theta \]

So:
\[\large \underbrace{2(3) + (-4)(-8)}_{\text{using (1)}}=\underbrace{\sqrt{2^2+(-4)^2}\sqrt{3^2+(-8)^2}\cos \theta}_{\text{using (2)}}\\
\large 38 = \sqrt{20}\sqrt{73}\cos \theta\\
\large 38 = 2\sqrt{5}\sqrt{73} \cos \theta \\
\large 38 = 2 \sqrt{365} \cos \theta \\
\large \cos \theta = \frac{38}{2 \sqrt{365}}= \frac{19}{\sqrt{365}} \\
\large \theta = \arccos \left(\frac{19}{\sqrt{365}} \right) =0.105 \text{ rad}=6.009^\text{o} \]