Find the angle between the given vectors to the nearest tenth of a degree. u = , v =

Question

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

amistre64 · Answer

have you learned about dot product of vectors?

amistre64 · Answer

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

brydenwright · Answer

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°

amistre64 · Answer

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

brydenwright · Answer

$$\sqrt{37} and \sqrt{65}$$

amistre64 · Answer

good so sqrt 2405 gives us ...49.0408

amistre64 · Answer

and the dot of the vectors?

brydenwright · Answer

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

amistre64 · Answer

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

amistre64 · Answer

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

brydenwright · Answer

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

amistre64 · Answer

good luck :)