Question

Find the angle between the given vectors to the nearest tenth of a degree.
u = <6, 4>, v = <7, 5>

Answer 1

familiar with dot product theorem?

Answer 2

\(<6,4>\cdot<7,5>= |u||v|cos(\theta)\)
so
\(6*7+4*5 = |u||v|cos(\theta)\)
so
\(\frac{62}{|u||v|}=\cos(\theta)\)

Answer 3

would the answer be 11.8 degrees?

Answer 4

\(|u|=\sqrt{6^2+4^2}=\sqrt{44}\)
\(|v|=\sqrt{7^2+5^2}=\sqrt{74}\)

Answer 5

\(\cos(\theta)=\frac{62}{\sqrt{44}\sqrt{74}}\)
so
\(\theta=\cos^{-1}(\frac{62}{\sqrt{44*74}}) \approx0.413\)

Answer 6

That's not one of my answer choices.

1.8°
-9.1°
0.9°
11.8°

Answer 7

@zzr0ck3r

Answer 8

woops \(\sqrt{6^2+4^2}=\sqrt{52}\)

Answer 9

so 1.8 degrees

Answer 10

\(\theta=\cos^{-1}(\frac{62}{\sqrt{52*74}}) \approx 0.0322\approx1.8^{\circ} \)

Answer 11

wonderful, thanks!

Answer 12

np