Question

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

u = <2, -4>, v = <3, -8>

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

Is it neither?

Answer 3

\(\huge ?\)

Answer 4

my bad I meant 6.0 degrees?

Answer 5

hmm what did you get for the dot product?

Answer 6

-4(-8)

Answer 7

\(\bf <a,{\color{blue}{ b}}>\cdot <c,{\color{blue}{ d}}>\implies (a\cdot c)+({\color{blue}{ b}}\cdot {\color{blue}{ d}})\)

Answer 8

\(\bf <a,b>\qquad magnitude\implies \sqrt{a^2+b^2}\)

Answer 9

So what would the answer come out to be?

Answer 10

well. divide the dot product by the magnitudes product
then get the \(\bf cos^{-1}\)

Answer 11

3.0 degrees?

Answer 12

hmm dunno

Answer 13

k, I'm pretty sure my first answer was correct. Thanks for the help

Answer 14

yw