## anonymous one year ago 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|

1. zzr0ck3r

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.

2. zzr0ck3r

Can you do this?

3. anonymous

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

4. zzr0ck3r

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

5. zzr0ck3r

$$||\text{v}_2||=\sqrt{(-3)^2+6^2}=\sqrt{54}$$ $$||\textbf{v}_1||\times ||\textbf{v}_2||=2(\sqrt{13})(\sqrt{54})$$

6. zzr0ck3r

I said something wrong, it should be $$\cos(\theta) =\dfrac{\mathbf{v}_1\circ \mathbf{v}_2}{||\mathbf{v}_1||\times ||\mathbf{v}_2||}$$

7. zzr0ck3r

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 $$\approx 37.57^{\circ}$$

8. anonymous

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?

9. zzr0ck3r

the numerator was the dot product