Question

help @vocaloid

Answer 1

I'm almost done with this lesson I promise

Answer 2

well you would copy down the formula from before, cos(theta) = u dot v/|u| |v| then explain that you would need to calculate the dot product for the numerator, the magnitudes for the denominator, solve for theta, etc.

Answer 3

well I guess the first step would be reading the graph and determining what the vectors for u and v would be in vector notation, then followed by those other steps

Answer 4

could we do the graph like work it out?

Answer 5

since both the vectors are at the origin, you just need to find the coordinates of the end points and then convert those to a vector to get your vector notation

for example, the endpoint of v is (-2,-1) so the vector is <-2,-1> then repeat the same logic for vector u

Answer 6

-3,-4

Answer 7

good, that's vector u, so vector v = <-2,-1> and u = <-3,-4> then you would calculate cos(theta) = u dot v/|u| |v| based on this to find theta

Answer 8

I'm not sure if they want the actual numbers and calculations or if they just want the steps but that's how you'd do it

Answer 9

still there? is there anything that's still unclear?

Answer 10

yes and hold on

Answer 11

so would I do the multiplication and squared thing?

Answer 12

@Vocaloid

Answer 13

I'm not sure what you mean, but if you're talking about magnitude and dot product calculations then yes

Answer 14

ok so i did all that and got 4^2+6^2 =52 and then root and got 7.21

Answer 15

@Vocaloid

Answer 16

try to keep the magnitude calculations and the dot product calculations separate

first, do the numerator calculation by
1. multiplying the x-coordinates
2. multiplying the y-coordinates
3. adding the answers from steps 1 and 2

Answer 17

got that

Answer 18

then, you would calculate the magnitude of u, magnitude of v and then calculate the denominator based on that

Answer 19

so 4^2 and 6^2

Answer 20

wouldn't I add those and root them?

Answer 21

nope, you would calculate u and v separately because the formula calls for the magnitudes of u and v separately

Answer 22

start with vector u, find the magnitude of <-2,-1> by taking r = sqrt(x^2 + y^2) and then move onto vector v

Answer 23

-5 for u

Answer 24

-2.24 for v

Answer 25

keep in mind magnitudes have to be positive.

sqrt((-2)^2 + (-1)^2)) = ?
sqrt((-3)^2 + (-4)^2)) = ?

Answer 26

5
25

Answer 27

don't forget the sqrts, so | u | = sqrt(5) and | v | = 5

then plug these into the denominator of the formula cos(theta) = u dot v/|u| |v|

once you plug in the dot product of u and v then you can solve for theta

Answer 28

so i got 5 and 2.24

Answer 29

so how would I set it up exactly?

Answer 30

\(\color{#0cbb34}{\text{Originally Posted by}}\) @Vocaloid
don't forget the sqrts, *** so | u | = sqrt(5) and | v | = 5 ***

then plug these into the denominator of the formula cos(theta) = u dot v/|u| |v|

once you plug in the dot product of u and v then you can solve for theta
\(\color{#0cbb34}{\text{End of Quote}}\)

Answer 31

I got the square root part 25=5 5=2.24

Answer 32

we have just determined that |u| = sqrt(5) and |v| = 5

the formula cos(theta) = u dot v/|u| |v| gives us the angle for u and v

simply plug in the magnitudes u and v, plug in the dot product, and calculate theta.

Answer 33

what was the dot product again I have steps down but it's kinda all over the place

Answer 34

\(\color{#0cbb34}{\text{Originally Posted by}}\) @Vocaloid
try to keep the magnitude calculations and the dot product calculations separate

first, do the numerator calculation by
1. multiplying the x-coordinates
2. multiplying the y-coordinates
3. adding the answers from steps 1 and 2
\(\color{#0cbb34}{\text{End of Quote}}\)

Answer 35

^ that's your dot product

Answer 36

so do i do the ^2 in dot product trying to not confuse myself

Answer 37

the dot product being 52 square root ?

Answer 38

the dot product does not involve squaring

simply identify the x-coordinates of u and v, the y-coordinates of u and v, and calculate the dot product

Answer 39

ok and i just add them no squaring or root or anything?

Answer 40

no squaring or rooting, no

Answer 41

so then just 10

Answer 42

yeah good, then plug everything in and solve for theta

Answer 43

anyway I really need to take a shower and go to sleep

cos(theta) = 10/(5sqrt(5))

theta = arccos(10/5sqrt(5)) then just chuck this into a calculator to get theta

Answer 44

you don't have time for 2 more questions?