Question

Please Help!!

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

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

Answer 1

@zzr0ck3r @Venny

Answer 2

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

Answer 3

So how would I wright that with the numbers? @jdoe0001

Answer 4

find the dot product... you should have covered that
and then find the magnitude

Answer 5

check your material on the dot product of two vectors and their magnitude, it should be there

Answer 6

(-5-4)*(-4-3) is the dot product?

Answer 7

\(\large <a,b>\cdot <c,d>\implies a\cdot c+b\cdot d\)

Answer 8

so (-5*-4)+(-4*-3)

Answer 9

yeap

Answer 10

how do i find the magnitude?

Answer 11

\(\large ||<a,b>||\implies \sqrt{a^2+b^2}\)

Answer 12

(6.4)=u
(5.3)=v??

Answer 13

I'm not understanding this @jdoe0001

Answer 14

hmm

Answer 15

\(\bf u = <-5, -4>, v = <-4, -3>
\\ \quad \\

\theta = cos^{-1}\left(\cfrac{u \cdot v}{||u||\ ||v||}\right)
\\ \quad \\
\theta = cos^{-1}\left(\cfrac{(-5\cdot -4)+(-4\cdot -3)}{\sqrt{(-5)^2+(-4)^2}\cdot \sqrt{(-4)^2+(-3)^2}}\right)\)

Answer 16

that will give you a value, that will be \(\bf -1 \ge\ or \le 1\)

Answer 17

ok thank you!! i got the answer!!

Answer 18

yw