Question

Please help, I don't understand at all. Let v1 = (-6, 4) and v2 = (-3, 6). Compute the following.
v1*v2

The angle between |v1| and |v2|

Answer 1

The angle \(\alpha\) between \(\mathbf{v}_1\text{ and } \mathbf{v}_2\) is given by \[\alpha=\dfrac{\mathbf{v}_1\circ \mathbf{v}_2}{||\mathbf{v}_1||\times ||\mathbf{v}_2||}\]

Where \(\circ\) is the dot product, and \(\times\) is normal multiplication on the reals.

Answer 2

Can you do this?

Answer 3

So the dot product should be (-6*-3)+(4*6) so 18+24 or 42.
What would the multiplication on the reals be?

Answer 4

\(||\textbf{v}_1||= \sqrt{(-6)^2+4^2}=\sqrt{52}=2\sqrt{13}\)

Answer 5

\(||\text{v}_2||=\sqrt{(-3)^2+6^2}=\sqrt{54}\)


\(||\textbf{v}_1||\times ||\textbf{v}_2||=2(\sqrt{13})(\sqrt{54})\)

Answer 6

I said something wrong, it should be

\(\cos(\theta) =\dfrac{\mathbf{v}_1\circ \mathbf{v}_2}{||\mathbf{v}_1||\times ||\mathbf{v}_2||}\)

Answer 7

So we have \(\cos(\theta) = \dfrac{42}{2\sqrt{13}\sqrt{54}}\implies \theta = \arccos(\dfrac{42}{2\sqrt{13}\sqrt{54}})\approx 0.65574\) in radians

Which is \(\approx 37.57^{\circ}\)

Answer 8

Ah, thank you! I was getting stuck because my textbook erroneously said the numerator wouldn't be the dot product. So I'm assuming v1*v2 would just be the dot product too?

Answer 9

the numerator was the dot product