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

racnna
Messages
40
Reaction score
0

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.
 
Physics news on Phys.org
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:
hmm...it does work...

ok i need to ask a follow up question but i'll create a new thread
 
If \alpha, \beta, and \gamma are three angles, the unit vector that makes angle \alpha with the x-axis, angle \beta with the y-axis and \gamma with the z-axis is cos(\alpha)\vec{i}+ cos(\beta)\vec{j}+ cos(\gamma)\vec{k}. 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 \sqrt{x^2+ x^2+ x^2}= x\sqrt{3} and we want that equal to 1: we want x= 1/\sqrt{3}= \sqrt{3}/3. The unit vector that makes equal angles with the coordinate axes is (\sqrt{3}/3, \sqrt{3}/3, \sqrt{3}/3).
 
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
 
Thread 'Use greedy vertex coloring algorithm to prove the upper bound of χ'
Hi! I am struggling with the exercise I mentioned under "Homework statement". The exercise is about a specific "greedy vertex coloring algorithm". One definition (which matches what my book uses) can be found here: https://people.cs.uchicago.edu/~laci/HANDOUTS/greedycoloring.pdf Here is also a screenshot of the relevant parts of the linked PDF, i.e. the def. of the algorithm: Sadly I don't have much to show as far as a solution attempt goes, as I am stuck on how to proceed. I thought...
Back
Top