Question

Find the angle θ between the vectors. (Round your answer to two decimal places.)
u = 20i + 40j
v = −3j + 9k
the answer i got is 31.95 but its wrong...

Answer 1

31.95 degree

Answer 2

What course are you in?

What vector operation are you using to obtain the cosine of the angle between the two vectors?

Answer 3

20 * -3 +40 * -3 then you take ||u|| * ||v|| then you simply use inverse of cos to get the answer

Answer 4

Thank you for sharing your result (31.95 degrees), but I'm far more interested in how you obtained that.

Answer 5

You're basically on the right track! But could you fix up the grammar and math involved in 20 * -3 +40 * -3 then you take ||u|| * ||v|| ?

What do you mean by "you take ... " ?

Answer 6

360/sqareroot of 2000 * 3 square root of 10 then just used the inverse of cos

Answer 7

Parth, supposing you were seeing the dot product for the first time, and trying to apply it. Would you, in that situation, likely understand "20 * -3 +40 * -3 then you take ||u|| * ||v|| "??

Answer 8

this is what i did first thing you have to find dot product of U dot V for dot product my result is 360.

now i have to find ||U|| the answer for that is Sqareroot of 2000 or 44.72

now i have to find ||V|| the answer for that is 3 square root of 10

then i did 360/squareroot of 2000 times 3 square root of 10

answer for the above steps = 0.848528137

then inverse of cos of 0.848528137

Answer 9

and thats how i got 31.95

Answer 10

What I'm asking you to do, Parth, is to type out the exact formula that we need here. It is\[\cos \theta=\frac{ u~\dot~v }{ |u||v| }\] I have trouble finding DIVISION within your explanation.

Answer 11

yes thats the one

Answer 12

sorry i don't really know how to put formula like that

Answer 13

i just joined today so

Answer 14

Therefore, you'd have this result:\[\cos \theta=\frac{ u~\dot~v }{ |u||v| }=\frac{ <20,40><-3,9> }{ |u||v| }\]

Answer 15

yes and i got 360 for the top part

Answer 16

Welcome aboard OpenStudy!!!
Have you already done the dot product indicated in the numerator? If so, what is that dot product?

Answer 17

How did you get 360? My result is -60+360, or ... ?

Answer 18

wait it should be 300 sorry

Answer 19

Much better. So, now, Parth, you have the scalar quantity 300 in the numerator.
What is the magnitude of vector u, please?

Answer 20

the answer for the question should be 45 degree?

Answer 21

square root of 2000

Answer 22

and for V = 3 square root of 10

Answer 23

Yes, and what is the magnitude of vector v?

Answer 24

\[3\sqrt{10}\]

Answer 25

45 degree is wrong answer

Answer 26

Looks better now. Multiply those two magnitudes together, please. Let's not focus on the answer yet, ok, but rather focus on how we get there accurately.

Answer 27

424.2640687

Answer 28

thank you for taking your time to help!

Answer 29

I'm going to take your word for that now, since it looks like it's in the correct ball park!
So, the cosine of your angle is 300 over 424.3, right? What is the angle in degrees? In radians?

Answer 30

Are you located in India? If so, in which city?

Answer 31

degree

Answer 32

no, i am in Texas, USA

Answer 33

Parth, let's try those absolute values once more.

Answer 34

You got a denom. of 424 something, whereas mine was 684 something. We have to be 100% in agreement on this.

vector u is <40,60>, right? double check the original problem statement.

Answer 35

u = 20i + 40j so it should be <20, 40>

Answer 36

The square of 40 is 1600 and that of 60 is 3600, correct?
Adding those squares together, we get 5200, correct?

Answer 37

If so, just find the amplitude of u as a mixed decimal number.

Answer 38

For vector v: the magnitude is the square root of 9+81, or square root of 90, or 3Sqrt(10). I'm in agreement with you there.

Answer 39

yeah but for U i have <20, 40> so 20^2 + 40^2 = square root of 2000

Answer 40

Yes, that looks correct, so the denominator is Sqrt(2000) time 3 times Sqrt (10).

Does that boil down to about 424?
If so, you're on the right track and I'm wrong.

Answer 41

yes 424.26

Answer 42

so the cosine of your angle is 300/424.26. What is your angle, in degrees?

Answer 43

300/424.26 =0.707113562

Answer 44

then inverse of cos (0.707113562) = 45 degree

Answer 45

Right!! Great work.

My advice: slow down a little; check everything twice.

Answer 46

but the answer is wrong LMAO

Answer 47

its online homework and i tried 45 and it gave me wrong answer

Answer 48

What are your answer choices?

Are they expressed in degrees or in radians?

Answer 49

i don't have answer choice but its asking in degree

Answer 50

(Thinking).

Answer 51

just hold on me take a screenshot and post it here

Answer 52

Hey, could you possibly take a screen shot of the original problem and share the image with me thru openstudy?

Answer 53

Great minds think alike!!

Answer 54

haha yeah there you go

Answer 55

see my first answer was 45 but i guess its wrong

Answer 56

I've done a quick, rough sketch of these two vectors; from that sketch it does seem that the angle is in the ballpark of 45 degrees. Let me look at your screenshot.

Answer 57

Thank you so much

Answer 58

Wow. As you know, "a picture is worth a thousand words."

Answer 59

The reason why we're getting the wrong answer is that these vectors are in 3-d space, not in the plane. the first one is <20,40,0> and the second is <0,-3,9>.
Are you able to find the dot product when you have vectors in 3-d space?

Answer 60

uhh.. idk where did i go worng!

Answer 61

Try finding the dot product <20,40,0> . <0,-3,9>.

Answer 62

OHHHHHHHHHHHHH

Answer 63

yeah its j, h, k thing

Answer 64

We have the magnitudes of the vectors correct, so try letting cos theta =-120/424 something.

Answer 65

i mean I J K

Answer 66

Yes, a i, j, k thing, otherwise known as 3-d space! ;)

Answer 67

the answer should be 106.44?

Answer 68

LOL!! thank you so much, i didn't look at the alphabets hahah

Answer 69

That's exactly what I got. Congrats!!!!

Answer 70

You are hereby condemned to look at the alphabet all night.

Answer 71

let me try that answer

Answer 72

Always a good idea to check. Curious, though, just how you'll check that answer.

Answer 73

its wrong and i am out of tries now so i have to talk with the professor to get it fixed again. thank you so much!!

Answer 74

I think you've learned a lot by going thru this. I'd suggest you try drawing these 2 vectors in space and see whether that gives you a clue as to what the angle should be.

Answer 75

yeah thank you!

Answer 76

By all means please come back to OpenStudy. I look forward to working with you again!
Good night!