Question

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

u = <-5, 8>, v = <-4, 8>

Answer 1

For any two vectors \(\mathbf{u}\) and \(\mathbf{v}\), it is true that
\[\cos\theta=\frac{\mathbf{u} \cdot\mathbf{v}}{|\mathbf{u}||\mathbf{v}|}\]
To find the angle, solve for theta.

Answer 2

I've done that, but every time I get to cos theta= 84/√41√128, I get stuck.

Answer 3

I will come to the same result as @blockcolder but with a derivation. If we look at the dot product of the two vectors then we get:\[\bf \vec{u} \frac{}{} \vec{v}=|\vec{u}||\vec{v}|\cos \theta\]And \(\bf \theta\) is the angle between the two vectors. Hence re-arranging the dot product yields:\[\bf \implies \cos \theta=\frac{ \vec{u} \frac{}{} \vec{v} }{ |\vec{u}||\vec{v}| }\]

Answer 4

I still don't really understand :(.

Answer 5

Ok first evaluate \(\bf \vec{u} \frac{}{} \vec{v}\):\[\bf =-5(-4)+8(8)=20+64=84\]Now let' s get their magnitudes and multiply them:\[\bf |\vec{u}|=\sqrt{(-5)^2+(8)^2}=\sqrt{89}\]\[\bf |\vec{v}|=\sqrt{(-4)^2+(8)^2}=\sqrt{16+64}=\sqrt{80}=4\sqrt{5}\]Hence:\[\bf \cos \theta=\frac{ 84 }{ \sqrt{89}(4\sqrt{5}) }=\frac{ 21 }{ \sqrt{445} }\]Now take the i nverse cosine:\[\bf \implies \theta=\cos^{-1}\left( \frac{ 21 }{ \sqrt{445} } \right) \approx 5.44°\]

Answer 6

@anbrii you understand?

Answer 7

Yes!! Thank you so much! I was finding the ||u|| ||v|| incorrectly. Instead of finding the square root of -4^2 plus 8^2, I was finding the squrt of 8^2 plus 8^2. Silly mistake. Thanks, again!