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

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SUMMARY

The discussion centers on finding a vector that forms equal angles with the x, y, and z axes. The initial assumption that the vector (1, 1, 1) meets this criterion is clarified as incorrect. The correct unit vector that makes equal angles with all three axes is determined to be (\sqrt{3}/3, \sqrt{3}/3, \sqrt{3}/3). This conclusion is reached by recognizing that the angles with the axes are represented by the cosines of the angles, which are equal when the vector components are the same.

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Homework Statement

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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.
 
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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:
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
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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