anonymous one year ago Find the angle between the given vectors to the nearest tenth of a degree. u = <6, -1>, v = <7, -4>

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1. Michele_Laino

$\Large \cos \theta = \frac{{u \cdot v}}{{\left| u \right|\left| v \right|}}$

2. Michele_Laino

where u.v is the scalar product between u and v, furthermore |u| is the length of the vector u, similarly for |v|

3. Michele_Laino

hint: $\Large \begin{gathered} \left| u \right| = \sqrt {{6^2} + {{\left( { - 1} \right)}^2}} = ... \hfill \\ \left| v \right| = \sqrt {{7^2} + {{\left( { - 4} \right)}^2}} = ... \hfill \\ u \cdot v = 6 \times 7 + \left( { - 1} \right) \times \left( { - 4} \right) = ... \hfill \\ \end{gathered}$

4. anonymous

So 0.2º?

5. Michele_Laino

I got this: $\Large \cos \theta = \frac{{u \cdot v}}{{\left| u \right|\left| v \right|}} = \frac{{46}}{{\sqrt {37} \sqrt {65} }} \cong 20.3{\text{degrees}}$

6. Michele_Laino

oops.. $\Large \cos \theta = \frac{{u \cdot v}}{{\left| u \right|\left| v \right|}} = \frac{{46}}{{\sqrt {37} \sqrt {65} }} \cong 0.93799$

7. Michele_Laino

so: $\Large \theta = 20.3\;{\text{degrees}}$

8. anonymous

Thanks!

9. Michele_Laino

:)

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