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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
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

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zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0The 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So the dot product should be (6*3)+(4*6) so 18+24 or 42. What would the multiplication on the reals be?

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0\(\textbf{v}_1= \sqrt{(6)^2+4^2}=\sqrt{52}=2\sqrt{13}\)

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0\(\text{v}_2=\sqrt{(3)^2+6^2}=\sqrt{54}\) \(\textbf{v}_1\times \textbf{v}_2=2(\sqrt{13})(\sqrt{54})\)

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0I said something wrong, it should be \(\cos(\theta) =\dfrac{\mathbf{v}_1\circ \mathbf{v}_2}{\mathbf{v}_1\times \mathbf{v}_2}\)

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0So 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 \(\approx 37.57^{\circ}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ah, 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?

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0the numerator was the dot product
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