Question

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

u = <-5, 8>, v = <-4, 8>

2.7°

-7.3°

5.4°

15.4°

Answer 1

\[\cos \theta = \frac{ u.v }{ \left| u \right|\left| v \right| }\]

Answer 2

the cosine of the angle equals the dot product of the two vectors divided by the product of the magnitudes of the two vectors

Answer 3

you need to find 3 things:
the dot product
the "length" of vector u
the "length" of vector v

can you do that ?

Answer 4

no i dont knowthe formulas for those

Answer 5

you know how to do those, because you did it in the last problem.

Answer 6

oh ok onesec

Answer 7

-5*-4+8*8 =84

Answer 8

yes that is the dot product.

what is the length of each vector ?
http://www.dummies.com/how-to/content/calculating-magnitude-with-vectors.html

Answer 9

sqrt(-5^2+-4^2) = i sqrt(41)
sqrt(8^2+8^2) = 8 sqrt(2)

Answer 10

try again. First, you use the components of u to find | u |
so it is sqrt( (-5)^2 + 8^2 )

also, it is not -(5^2) it is (-5)^2

Answer 11

sqrt( (-5)^2 + 8^2 ) =89^(1/2)
sqrt( (-4)^2 + 8^2 ) = 4 sqrt(5)

Answer 12

yes, but I would change them to decimals (keep at least 3 decimals)
because the question asks for nearest tenth of a degree.

now multiply your lengths you get (about) 9.434 * 8.944= 84.38

now find u dot v / 84.38

Answer 13

you figured out u dot v up above. it is 84

Answer 14

whats
u dot v

Answer 15

84/ 84.38

Answer 16

\[ u \cdot v \] is the dot product of u and v

you started with the formula
\[ \cos x = \frac{ u \cdot v}{|u| \ |v| } \\ \cos x= \frac{84}{84.38}\\ x= \cos^{-1} \left( \frac{84}{84.38}\right)\]

can you find x (the angle between the vector u and vector v ) ?

Answer 17

0.9954965631666271628347949751125859208343209291301256

Answer 18

you want the inverse cosine of that number

Answer 19

you can type
acos(0.995496563) in degrees=
into the google search window (or use a calculator)

Answer 20

5.4 is the answer thx man