Question

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

u = <6, -1>, v = <7, -4>

20.3°
10.2°
0.2°
30.3°

Answer 1

@ThisGirlPretty

Answer 2

Welcome to questioncove! and I'm sorry, I'm not good in math v.v @Nnesha HELPPPP @Vocaloid @Hero

Answer 3

@Nnesha

Answer 4

Is it 10.2?

Answer 5

here is the formula u use to find the angle \[\huge\rm \cos \theta = \frac{u \cdot v}{\left| u \right|\left| v \right| } \]

Answer 6

so i just plug in 6,1 and 7,-4 in u and v and thats the answer?

Answer 7

i didn't get 10.2. can you post your work please?

Answer 8

well first you have to find the dot product of u/v
and then find the magnitude of each vector
are you familiar with dot product and magnitude ?

Answer 9

No , i just started this lesson and im very confused

Answer 10

i think i know what the dot product is

Answer 11

ok
dot product is a method to multiply the vectors
if the given vectors are \[\large\rm u= < a_1 , b_2> ~~~,~~v= <a_2,b_2>\]

Answer 12

i think its 46 , because 6*7 + -1* -4

Answer 13

ok
dot product is a method to multiply the vectors
if the given vectors are \[\large\rm u= < a_1 , b_2> ~~~,~~v= <a_2,b_2>\]
\

Answer 14

yes that's right

Answer 15

\(\color{#0cbb34}{\text{Originally Posted by}}\) @Nnesha
ok
dot product is a method to multiply the vectors
if the given vectors are \[\large\rm u= < a_1 , b_2> ~~~,~~v= <a_2,b_2>\]
\(\color{#0cbb34}{\text{End of Quote}}\)
this is what i was trying to post https://prnt.sc/g418y6

Answer 16

and the magnitude will be \[\left| u \right| = \sqrt{a^2+b^2}\]

Answer 17

i'm confused , what numbers do i use to find the magnitude

Answer 18

there are different ways to write a vector
\[\large\rm u= \color{Red}{a}i +\color{blue}{b}j ~~~~or~~~~ u= <\color{red}{a},\color{blue}{ b}>\]
\

Answer 19

\(\color{#0cbb34}{\text{Originally Posted by}}\) @Nnesha
there are different ways to write a vector
\[\large\rm u= \color{Red}{a}i +\color{blue}{b}j ~~~~or~~~~ u= <\color{red}{a},\color{blue}{ b}>\]
\
\(\color{#0cbb34}{\text{End of Quote}}\)
https://prnt.sc/g41cen

Answer 20

so the magnitude of u vector is \[\ \left| u \right|= \sqrt{a^2+b^2} = \sqrt{(6)^2+(-1)^2}\]

Answer 21

\[\sqrt{37}\]

Answer 22

so the magnitude is square root 37?

Answer 23

do i do the same for v?

Answer 24

yes right and yes

Answer 25

\[\sqrt{(7)^2+(3)^2 } = \sqrt{58}\]

Answer 26

it's -4 not 3 so that should be sqrt{65}

Answer 27

oh your right

Answer 28

what do i do with \[\sqrt{37} \sqrt{65}\]

Answer 29

\(\color{#0cbb34}{\text{Originally Posted by}}\) @Nnesha
here is the formula u use to find the angle \[\huge\rm \cos \theta = \frac{u \cdot v}{\left| u \right|\left| v \right| } \]
\(\color{#0cbb34}{\text{End of Quote}}\)
now u can plugin the values
the numerator (dot product ) = 46
denominator (magnitude of u and v) = sqrt{65} times sqrt{37}

Answer 30

0.93799440

Answer 31

that's what i got when i plugged it into my calculator

Answer 32

\[\cos \theta = 0.93799440
\]
take inverse cos to find theta

Answer 33

\[\theta= 0.591406\]

Answer 34

\[\large\rm \theta = \cos^{-1} (0.93799440)\]

Answer 35

use the calculator^

Answer 36

\[\theta= 0.353998\]

Answer 37

looks right

Answer 38

But thats not one of the answers

Answer 39

do you have the answer choices ??
maybe they want the answer in degree not radians ?

Answer 40

20.3°
10.2°
0.2°
30.3°

Answer 41

ohh okay so they want the answer in degree not radians
in order to convert the answer to degree multiply it by 180/pi

Answer 42

20.282591356

Answer 43

when rounded its A , correct?

Answer 44

yes

Answer 45

THANK YOU SO MUCH

Answer 46

yw :=)) o^_^o

Answer 47

Good job Nnesha and word2 :)!!