Question

Find the angle between the given vectors to the nearest tenth of a degree. u = <-5, -4>, v = <-4, -3>

Answer 1

\(\bf \textit{angle between two vectors }\\ \quad \\
cos(\theta)=\cfrac{u \cdot v}{||u||\ ||v||} \implies \cfrac{\text{dot product}}{\text{product of magnitudes}}\\ \quad \\
\theta = cos^{-1}\left(\cfrac{u \cdot v}{||u||\ ||v||}\right)\)

Answer 2

How do i find u and v?? would I use absolute value?? This is for my online class and im totally lost in this module

Answer 3

\(\Large u\) = <-5, -4>, \(\Large v \)= <-4, -3>

Answer 4

im so lost

Answer 5

well... if you have not covered the material
seems this exercise does not apply to you then

Answer 6

no I have but im trying to teach myslef since its online and I cant figure it out

Answer 7

The "dot product"

\[\large u \cdot v = |u||v|cos(\theta) \]

We want to solve for the angle...so we can divide both sides by \(\large |u||v|\)

\[\large cos(\theta) = \frac{u\cdot v}{|u||v|}\]

And now take the inverse cos of those sides

\[\large \theta = \cos^{-1}(\frac{u \cdot v}{|u||v|})\]

The \(\large |u|\) and \(\large |v|\) are the "magnitudes of the vectors"

\[\large |u| = \sqrt{u_x ^2 + u_y^2} \]
\[\large |v| = \sqrt{v_x ^2 + v_y^2} \]

Just as the dot product is calculated as

\[\large u \cdot v = (u_x \times v_x + u_y \times v_y)\]

That should be everything you need to do this problem! Good luck in your self-studying