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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|

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  1. zzr0ck3r
    • one year ago
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    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
    • one year ago
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    Can you do this?

  3. anonymous
    • one year ago
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    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
    • one year ago
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    \(||\textbf{v}_1||= \sqrt{(-6)^2+4^2}=\sqrt{52}=2\sqrt{13}\)

  5. zzr0ck3r
    • one year ago
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    \(||\text{v}_2||=\sqrt{(-3)^2+6^2}=\sqrt{54}\) \(||\textbf{v}_1||\times ||\textbf{v}_2||=2(\sqrt{13})(\sqrt{54})\)

  6. zzr0ck3r
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    the numerator was the dot product

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