Question

Find the angle between the given vectors to the nearest tenth of a degree.
u = <6, -1>, v = <7, -4>

Answer 1

for any 2 vectors, the following relationship holds
\[\cos \theta = \frac{u*v}{|u|*|v|}\]

Answer 2

so how would i proceed with this problem?

Answer 3

find magnitudes of u and v , find dot product of u and v
then solve for theta

Answer 4

could you please show me how to do that step by step? :)

Answer 5

dot product of 2 vectors
\[u = <a,b> , v=<c,d>\]
\[u*v = (a*c)+(b*d)\]

magnitude of a given vector
\[u = <a,b>\]
\[|u| = \sqrt{a^{2} +b^{2}}\]

Answer 6

i got 8.06 and 6.08

Answer 7

for what ?

Answer 8

the magintude of the vectors?

Answer 9

ahh yes correct....i left them as sqrt(37) and sqrt(65)

Answer 10

be careful with rounding since that can affect the final answer

Answer 11

what is the angle between them? :)

Answer 12

ok what is dot product (u*v) ?

Answer 13

42

Answer 14

not quite...should get 46

--> 7*6 + (-1)*(-4) = 42+4 = 46

Answer 15

\[\cos \theta = \frac{46}{\sqrt{37} \sqrt{65}}\]
take inverse cos (use calculator)
\[\theta = \cos^{-1} \frac{46}{\sqrt{37} \sqrt{65}} \approx 20.3^{o}\]

Answer 16

thank you :)