Question

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

Answer 1

have you learned about dot product of vectors?

Answer 2

\[|\vec u|~|\vec v|~cos(\alpha)=\vec u \cdot \vec v \]
and solve for \(\alpha \)

Answer 3

I tried that, but I got 133.89 and the answer choices are
a. 20.3°
b. 10.2°
c. 0.2°
d. 30.3°

Answer 4

lets see how well you calculated the parts ... what did you get for vector lengths?

Answer 5

\[\sqrt{37} and \sqrt{65}\]

Answer 6

good so sqrt 2405 gives us ...49.0408

Answer 7

and the dot of the vectors?

Answer 8

-1(6)+7(-4)=-34

Answer 9

<6, -1>
<7, -4>
42+4 = 46 is what i get for the dot

Answer 10

you multiply like parts, and add the results ... your process is a little off is all

Answer 11

Oh wow, I made a stupid mistake. Thanks, I can solve it from here!

Answer 12

good luck :)