Question

vectors

Answer 1

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

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

Answer 2

i am not sure how you were taught to do this, but one way is to first compute the "dot product"
do you know what that is?

Answer 3

i can show you if you like

Answer 4

no i dont know it .-.

Answer 5

\(<u_1,u_2>\cdot<v_1,v_2>=u_1v_1+u_2v_2\)

Answer 6

in your case it is \[(-4)(-5)+(-4)(3)=8\] put that in the numerator of a fraction

Answer 7

in the denominator goes the product of the norms of the vector
\[||<x,y>||=\sqrt{x^2+y^2}\]

Answer 8

in your case it is
\[\sqrt{4^2+5^2}\sqrt{3^2+4^2}=5\sqrt{41}\] if my arithmetic is correct

Answer 9

then to find the angle,
\[\cos(\theta)=\frac{8}{5\sqrt{41}}\] so
\[\theta =\cos^{-1}\left(\frac{8}{\sqrt{41}}\right)\]

Answer 10

since you are evidently working in degrees, put the calculator in degree mode, compute, and round to the nearest tenth as asked

Answer 11

i get about \(75.53\)
http://www.wolframalpha.com/input/?i=arcos%288%2F%285sqrt%2841%29%29%29

Answer 12

woah thats way too big.... the answers i can choose from are
-9.1°

1.8°

0.9°

11.8°

Answer 13

hmm let me see if i made a mistake, but i don't see it

Answer 14

oh damn, i do see it
it is \(-3\) not \(3\) so my dot product was wrong

Answer 15

20+12=32

Answer 16

that is better!

Answer 17

\[\cos^{-1}(\frac{32}{5\sqrt{41}})\] should work better

Answer 18

now i get \(1.79\) rounding to \(1.8\)