Question

Find the angle between the given vectors to the nearest tenth of a degree.

u = <6, -1>, v = <7, -4>

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

So 0.2º?

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

Thanks!

Answer 9

:)