Question

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

u = <-5, -4>, v = <-4, -3>

Answer 1

I am stuck at my last step please help me!

Answer 2

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

Answer 3

well let me show you where i got to
\[\frac{ (-5)(-4)+(-4)(-3) }{ \sqrt{-5^2+(-4)^2}\sqrt{-4^2+(-3)^2} }\]

Answer 4

then my answer was \[\frac{ 32 }{\sqrt{41}\sqrt{25} }\]

Answer 5

What do I do next?

Answer 6

\(\bf \frac{ (-5)(-4)+(-4)(-3) }{ \sqrt{(-5)^2+(-4)^2}\sqrt{(-4)^2+(-3)^2} }\qquad \huge \checkmark\)

Answer 7

\(\bf \textit{angle between two vectors }\\ \quad \\
cos(\theta)=\cfrac{u \cdot v}{||u||\ ||v||} \implies \cfrac{\text{dot product}}{\text{product of magnitudes}}\\ \quad \\
\theta = cos^{-1}\left(\cfrac{u \cdot v}{||u||\ ||v||}\right)\implies \theta=cos^{-1}\left(\cfrac{32}{\sqrt{41}\cdot \sqrt{25}}\right)\)

Answer 8

so do I have to use \[\cos^{-1} \frac{ 32 }{ \sqrt{41}\sqrt{25} }\]

Answer 9

hahah we replied at the same time!

Answer 10

hehe

Answer 11

so the degree is .9?

Answer 12

https://www.google.com/search?client=opera&rls=en&q=acos(32/(sqrt(41)*5))&sourceid=opera&ie=utf-8&oe=utf-8&channel=suggest#channel=suggest&q=acos(32%2F(sqrt(41)*5))+in+degrees&rls=en

Answer 13

\(\bf \theta=cos^{-1}\left(\cfrac{32}{\sqrt{41}\cdot \sqrt{25}}\right)\implies \theta=cos^{-1}\left(\cfrac{32}{\sqrt{41}\cdot 5}\right)\implies \theta \approx 1.7899^o\)

Answer 14

of I see i didn't put 5 there for some reason thanks!

Answer 15

yw