Calculate Magnitude & Angles of Vector B

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SUMMARY

The discussion focuses on calculating the magnitude and angles of Vector B with components (4, 6, 3). The magnitude is confirmed as 7.81. To find the angles with the coordinate axes, participants clarify that the angles can be derived using the formula arccos(component/magnitude), resulting in angles of arccos(4/7.81), arccos(6/7.81), and arccos(3/7.81). The conversation emphasizes the importance of visualizing the vector in a three-dimensional space, particularly in relation to a rectangular prism.

PREREQUISITES
  • Understanding of vector components in three-dimensional space
  • Familiarity with the concept of vector magnitude
  • Knowledge of trigonometric functions, specifically arccosine
  • Ability to visualize geometric shapes, such as rectangular prisms
NEXT STEPS
  • Study the properties of vectors in three-dimensional geometry
  • Learn about the dot product and its applications in finding angles between vectors
  • Explore the law of cosines and its proofs for triangle calculations
  • Practice problems involving vector magnitudes and angles in various contexts
USEFUL FOR

Students studying physics or mathematics, particularly those focusing on vector analysis and three-dimensional geometry. This discussion is beneficial for anyone seeking to understand vector magnitudes and angles in a spatial context.

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Homework Statement


Vector B has x,y,z components of 4,6 and 3 respectively. Calculate the magnitude of B and the angles B makes with the coordinate axes.


Homework Equations


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The Attempt at a Solution


I can do the magnitude part which is 7.81, but I don't really understand the angles part. Does it mean what angle it makes with the right triangle 5,6 and sqrt61? Please clarify and thanks!
 
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No, it's asking what angles does that vector make with the x, y, and z-axes. That is, what angle does B make with the vectors (1,0,0), (0,1,0), and (0,0,1).
 
Oh wow I'm dumb. Thanks!
 
Oh but I still don't get it. I know how you do the angles in 2-d planes, but how do you find it out in the x,y,z plane? I'm sorry, but I'm learning by myself and Chapter 3 does not say anything about angles in the x,y,z plane.
 
Do you know \mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}} <br /> \over A} \cdot \mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}} <br /> \over B} = \left| A \right|\left| B \right|\cos \theta? This will tell you the angle between any 2 vectors.
 
Never knew that. Why would the questions ask me that if they never taught it to me? Is this the dot product?
 
Also is there a proof, I can find? Or is it easy enough for me to prove?
 
Hi abhikesbhat :smile:

(4,6,3) is the diagonal of a 4,6,3 rectangular box …

they're asking the angle the diagonal makes with each of the three edges …

so draw the diagonals of the three faces which join at (4,6,3), and make three right-angled triangles: that will give you the angles :wink:
 
But how? Where it starts from is the bottom, but it has 3 components. Or is it asking the angle where the vector ends?
 
  • #10
abhikesbhat said:
But how? Where it starts from is the bottom, but it has 3 components. Or is it asking the angle where the vector ends?

It's asking for the angle at the origin between the diagonal and one edge …

so that's two sides of a triangle, and you just need to draw in the third side of the triangle, which will be along one of the far faces :wink:
 
  • #11
Ok so the right triangle on the xy plane is A=7.81 B=3 so aTan gives us an angle of 21. Is that right?
 
  • #12
Sorry for the impatience, but am I right?
 
  • #13
No, I don't think so... why are you using the arctangent?

Personally, I think the vector identity Pengwuino posted is the easiest way to do this problem. You can prove it using the law of cosines, and in turn several proofs of that law are listed on its Wikipedia page.
 
  • #14
abhikesbhat said:
Sorry for the impatience, but am I right?

erm :redface: … some of us have to sleep! :zzz:
abhikesbhat said:
Ok so the right triangle on the xy plane is A=7.81 B=3 so aTan gives us an angle of 21. Is that right?

No, A is the hypotenuse …

you need the arc-… ? :smile:
 
  • #15
abhikesbhat said:
Also is there a proof, I can find? Or is it easy enough for me to prove?

http://ocw.mit.edu/OcwWeb/Mathematics/18-02Fall-2007/VideoLectures/detail/embed01.htm

Maybe you'll find this helpful
 
  • #16
Ok tiny-tim I had some trouble visualizing it so used real life objects to model. The bottom face or the xz plane has a diagonal of 5 and the vector is 7.81. That makes a right triangle with hypotenuse 7.81 and the bottom side 5. aCos(5/7.81) is 50.19 degrees. Is that right? I'll be patient this time...
 
  • #17
Hi abhikesbhat! :smile:
abhikesbhat said:
Ok tiny-tim I had some trouble visualizing it so used real life objects to model. The bottom face or the xz plane has a diagonal of 5 and the vector is 7.81. That makes a right triangle with hypotenuse 7.81 and the bottom side 5. aCos(5/7.81) is 50.19 degrees. Is that right? I'll be patient this time...

Yes, the bottom face has sides of 3 and 4, so the diagonal is 5, and so the cosine is 5/7.81 …

but that's the angle between B and the xz plane …

you're asked for the angle between B and each of the three axes :wink:
 
  • #18
Hmm I don't get it still. Can show step by step how to get the answer if I get it wrong again? I have been stuck on this for awhile now. For the angle it makes with the x-axis, I got a triangle with hypotenuse 7.81 and bottom side 4. This is aCos(4/7.81), but I don't have a calc right now so the angle is aCos(4/7.81)?
 
  • #19
abhikesbhat said:
Hmm I don't get it still. Can show step by step how to get the answer if I get it wrong again? I have been stuck on this for awhile now. For the angle it makes with the x-axis, I got a triangle with hypotenuse 7.81 and bottom side 4. This is aCos(4/7.81), but I don't have a calc right now so the angle is aCos(4/7.81)?

Yes …

the three angles are arccos(4/7.81) arccos(6/7.81) and arccos(3/7.81) :smile:
 
  • #20
Oh thanks a bunch tiny-tim! I think I understand now, by picturing a rectangular prism. Thanks!
 

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