Question

help please! magnitude of V!

Answer 1

i need help with number 2 if possible

Answer 2

file won't open

Answer 3

The magnitude of vector \(\bf V\)
\[|{\bf V}|=V=\sqrt{V_x^2+V_y^2+V_z^2}\]

Answer 4

how do you know which one is which in order to solve?

Answer 5

unkle has the formula for magnitude right there

Answer 6

yeah i think i got that part but i really dont get c.) and d.)

Answer 7

sorry, been too long since this stuff, and we didn't get this deep into it.

Answer 8

what course is this from?

Answer 9

physics i believe

Answer 10

ahhh I haven't had that yet, I just got introduced into vectors in trig and a little in pre-calc

Answer 11

so confusing =(

Answer 12

I won't disagree

Answer 13

i am stuck on basically b, c, d... sorry =(

Answer 14

Do you know what projection of a vector means? Do you need an example to start with?

Answer 15

yes it would be great if i can get an example

Answer 16

Hmm, I'd draw it out. Could you wait for a while? I may even be shown offline.

Answer 17

k i will

Answer 18

\[\sin\theta=\frac{\text{opposite side}}{\text{hypotenus}}\]

Answer 19

@UnkleRhaukus it would be little bit easier if you shown me where and how to apply it.

Answer 20

http://imgur.com/FITxpgo

Answer 21

2 dimensions for starters.

Answer 22

how would you find the angle of theta y ?

Answer 23

and thank you i understood that picture!

Answer 24

find the angle of theta y, etc with trigonometry

Answer 25

Whoa, I just noticed that we have numbers there too.

Answer 26

yes haha but im still confused on finding the angle i mean using trig is the first step but can you be more specific? thank you =(

Answer 27

use the numbers,

Answer 28

Erm... what are those numbers pointing? Components of V or lengths of sides?

Answer 29

Those must be components of V. Part a) asks for the magnitude of V.

sin(theta_x) = 5 / |V|

Answer 30

If they are the components, we don't even need an angle. Life becomes easier.

Answer 31

True. But part b) asks for the angle theta_y and so for part c) theta_x can be found the same way.

Answer 32

Just remove the z-component and you get a vector in xy plane. This is also the vector which we otherwise would have calculated.

Answer 33

Uh, what is 3.5?

Answer 34

@ranga so i got magnitude to be 6.60 and then apply sin (theta) = 5/magnitude?

Answer 35

correct.

Answer 36

do you use inverse sin to find the angle?

Answer 37

yeah, inverse sine to find theta_x and then |V| * cos(theta_x) to find the component of V on the xy plane.

Answer 38

ok thank you! i will let you know!

Answer 39

And you can calculate the xy component another way and you can verify if you get the same answer. \[\Large V_{xy} = \sqrt{V_x^2 + V_y^2}\]

Answer 40

From the diagram, \[\Large V_x = 3.5 \quad V_y = 2.5 \quad V_z = 5.0\]

Answer 41

thank you so much ! that cleared things alot!

Answer 42

You are welcome.

Answer 43

Hey, do you need help for the fourth part?

Answer 44

oh yes please!

Answer 45

Sure. So theta x is the angle of V along x, theta y along y and theta z along z. Direction cosines are basically cos(theta x), cos(theta y) and cos(theta z).

Answer 46

oh i see so would you use the same method as top to find other angles also?

Answer 47

Probably not! As you see, cos(theta x) = Vx/|V|

Answer 48

think I made a mistake earlier in using theta_x. For c) simply use the square root formula rather than the angle.

Answer 49

@ranga yea i used the square root thanks!

Answer 50

You know the values of Vx, Vy, Vz and |V|.
cos(theta x) = Vx/|V|
cos(theta y) = Vy/|V|
cos(theta z) = Vz/|V|

Answer 51

@ParthKohli thanks you so much it makes more sense now! =)

Answer 52

Good job!

Answer 53

Do tell if you need any more help.

Answer 54

yea i would be in debt if you can help me with #3 part b.)

Answer 55

Oh my, I forgot about your post

Answer 56

Do you know how to resolve? First, pick suitable coordinate axes.

Answer 57

hm.. im not quite sure do you want to show me?

Answer 58

All right.

Answer 59

http://imgur.com/eqgAhdn

Answer 60

oh thanks what do you think about part b, c, d they all seem to be similar but i dont know how to start .. =( sorry

Answer 61

Do you know how to resolve OA?

Answer 62

i am not 100% sure sorry

Answer 63

i know it might be r0 - r A perhaps.

Answer 64

Do the same thing that I did in this pic. Use the angle and the vector, along with trigonometry.

Answer 65

Do you need me to draw it out?

Answer 66

oh yes please that would be most helpful ! @.@

Answer 67

http://imgur.com/I7UBpF6

Answer 68

Oh, you beat me to it!

Answer 69

Third part now?

Answer 70

oh wow im understanding alot yes 3rd part would be more helpful too xD

Answer 71

one question for part b.) is how on the picture how did u put 15 degree on the IV quadrant because pictures seems to be in the I quadrant ?

Answer 72

Do you know which vector r is in the picture?

Answer 73

http://imgur.com/yOluUm2