anonymous
  • anonymous
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|
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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zzr0ck3r
  • 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.
zzr0ck3r
  • zzr0ck3r
Can you do this?
anonymous
  • anonymous
So the dot product should be (-6*-3)+(4*6) so 18+24 or 42. What would the multiplication on the reals be?

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zzr0ck3r
  • zzr0ck3r
\(||\textbf{v}_1||= \sqrt{(-6)^2+4^2}=\sqrt{52}=2\sqrt{13}\)
zzr0ck3r
  • zzr0ck3r
\(||\text{v}_2||=\sqrt{(-3)^2+6^2}=\sqrt{54}\) \(||\textbf{v}_1||\times ||\textbf{v}_2||=2(\sqrt{13})(\sqrt{54})\)
zzr0ck3r
  • zzr0ck3r
I said something wrong, it should be \(\cos(\theta) =\dfrac{\mathbf{v}_1\circ \mathbf{v}_2}{||\mathbf{v}_1||\times ||\mathbf{v}_2||}\)
zzr0ck3r
  • 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 \(\approx 37.57^{\circ}\)
anonymous
  • 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?
zzr0ck3r
  • zzr0ck3r
the numerator was the dot product

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