Vector which has same angle with x,y,z axes

  • Thread starter racnna
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  • #1
racnna
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Homework Statement

?
How do i find a vector that has same angle with the three coordinate axes (x,y,z)?

The Attempt at a Solution



I immediately thought [1,1,1] would be it but it's not. I'm trying to find a plane whose normal vector forms the same angle with the three coordinate axes.
 

Answers and Replies

  • #2
Bacle2
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Why doesn't v=(1,1,1) work?

Let e,i,j stand for (1,0,0) ,(0,1,0) and (0,0,1) respectively.

Using a.b=|a||b|cost,

Then v.e=v.i=v.j =1 ; |e|=|i|=|j|=1 .Then

v.e/|v||e|= v.i/|v|.|i|= v.j/|v||j|=cost

Or did you have a different notion of angle in mind?
 
Last edited:
  • #3
racnna
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hmm...it does work...

ok i need to ask a follow up question but i'll create a new thread
 
  • #4
HallsofIvy
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If [itex]\alpha[/itex], [itex]\beta[/itex], and [itex]\gamma[/itex] are three angles, the unit vector that makes angle [itex]\alpha[/itex] with the x-axis, angle [itex]\beta[/itex] with the y-axis and [itex]\gamma[/itex] with the z-axis is [itex]cos(\alpha)\vec{i}+ cos(\beta)\vec{j}+ cos(\gamma)\vec{k}[/itex]. If all angles are the same, those three cosines are the same so any vector of the form (x, x, x), and in particular (1, 1, 1) will make equal angles with the three coordinate axes.

perhaps you are looking for the unit vector. The length of (x, x, x) is [itex]\sqrt{x^2+ x^2+ x^2}= x\sqrt{3}[/itex] and we want that equal to 1: we want [itex]x= 1/\sqrt{3}= \sqrt{3}/3[/itex]. The unit vector that makes equal angles with the coordinate axes is [itex](\sqrt{3}/3, \sqrt{3}/3, \sqrt{3}/3)[/itex].
 
  • #5
racnna
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yes i got the same result for the unit vector. Thanks Ivy. Please see my other thread on the stress tensor. its a follow up to this question
 

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