Vector of magnitude m making angles alpha, beta and gamma with i,j,k

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Main Question or Discussion Point

Many of us have seen how to find a vector satisfying the following conditions (i) magnitude is m (ii) makes angles alpha, beta and gamma with i,j,k vectors i.e. unit vectors along x,y and z axes.

My question is can the angles be taken arbitrarily or they should satisfy some condition like angles in a triangel add up to pi radians.

I have a dilemma. The method we see in books indicates like we can arbitrarily choose the angles.

But let us consider the following:

The vectors making gamma angle with z axix should necessarily be on a cone. Now the condition that it should make alpha with x axis restrains this vector space to vectors on the cone to a few that make alpha angle angel with x-aix. Out of these vectors are we always sure that there will be a vector which also makes beta angle with the y-axix, whatever be the alpha, beta and gamma?
 

Answers and Replies

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No we can not arbitrarily choose the angles. If you are given 2 angles, there are at most 2
possibilities left for the third. (intersection of two cones). the third angle is only given to tell wich of these 2 it ts.
 
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there is the equation about the 3 angle :cos(a)^2+cos(b)^2+cos(g)^2=1
if the alpha and beta are too small , their cosine will be too big to get the gamma
 
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But here is the dilemma. The following method looks logical and looks like yielding a vector whatever be the m, alpha, beta and gamma. Assume the desired vector to be d=ai+bj+ck. In that case, d.i = a = magnitude(d)..cosine(alpha). Since we know, magnitude of d (which is m) and alpha we can calculate a. Similarly we can calculate b and c and arrive at the desrired vector d whatever be m, alpha, beta and gamma. Where is the catch?
 
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Check that cos(alpha)= a/magnitude(d), which is determined by a ,b ,c together.
if you try to change a in order to change alpha, the magnitude will change with a,
so cos(beta) and cos(gamma) will change with a , since they also depends through
the magnitude(d). that is to say, beta and gamma is varying with alpha and hold a fix relation among the 3 angles, when you keep b, c constant and change a.
As a result, in effect you can't choose the angles arbitrarily
 
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Many of us have seen how to find a vector satisfying the following conditions (i) magnitude is m (ii) makes angles alpha, beta and gamma with i,j,k vectors i.e. unit vectors along x,y and z axes.
No. Since the magnitude is arbitrary only the direction needs to be determined. A direction requires two parameters. Let us use a vector of unit length as an example (since all other vectors with the same direction are merely a multiply of this unit vector). Once you specify the second angle there are only two choices for the third.
My question is can the angles be taken arbitrarily or they should satisfy some condition like angles in a triangel add up to pi radians.
There are only two choices for the third angle as far as I can recall.
I have a dilemma. The method we see in books indicates like we can arbitrarily choose the angles.
I doubt that is true. Can you show me a picture of what you're referring too?
But let us consider the following:

The vectors making gamma angle with z axix should necessarily be on a cone.
If the angle [itex]\gamma[/itex] is 90 degrees then it will lie in the xy-plane and not on a cone. In this case the second angle will then determine which direction the unit vector points in and that can be specified by one angle. As such the third angle will be uniquely determined. There are not two choices in this instance, only one.
Now the condition that it should make alpha with x axis restrains this vector space to vectors on the cone to a few that make alpha angle angel with x-aix.
Picture this in your mind; Choosing the first angle will determine a cone about the z-axis (unless the angle is 90 degrees). Choosing the angle the vector makes with the x-axis will determine a cone around the x-axis. Since the direction must satisfy the same angles then you're looking for the intersection of two cones. There are two directions specified by the interesection of these two cones.
Out of these vectors are we always sure that there will be a vector which also makes beta angle with the y-axis, whatever be the alpha, beta and gamma?
Once the first two angles are determine the second angle is restricted. If the first angle is not 90 degrees then there are two choices for the third. If the first angle is 90 degrees then once the second angle is chosen the third angle is uniquely determined.

Best wishes

Pete
 
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there is the equation about the 3 angle :cos(a)^2+cos(b)^2+cos(g)^2=1
if the alpha and beta are too small , their cosine will be too big to get the gamma
That expression can be solved for one of the angles when the other two are given. Since there will be two signs (one for the positive square root and one for the negative square root) unless it is zero. Therefore the last angle has at most two solutions. If one angle is 90 degrees then the second angle uniquely determines the direction and there is no other choice for the last, third, angle.

Pete
 
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But here is the dilemma. The following method looks logical and looks like yielding a vector whatever be the m, alpha, beta and gamma. Assume the desired vector to be d=ai+bj+ck. In that case, d.i = a = magnitude(d)..cosine(alpha). Since we know, magnitude of d (which is m) and alpha we can calculate a. Similarly we can calculate b and c and arrive at the desrired vector d whatever be m, alpha, beta and gamma. Where is the catch?
My question is where is the catch? I choose arbitrarily m, alpha, beta and gamma. I find a,b,c as mentioned above and as such arrive at a vector d which satisfies all the conditions of m, alpha, beta and gamma. Whatever be this four values. For example, I choose m =1000, alpha = pi/3, beta = pi/10, gamma = 3pi/5, I arrive at d. Even if some of the 3 angles are pi/2, pi or whatever.
 
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manjuvenamma:
this is a question about "necessary condition" and "sufficient condition."
if you have d=ai+bj+ck, you can get di=a by mutiplying i, similiarly dj=b and dk=c;
but if you only have di=a, dj=b,and dk=c, you CANNOT obtain d=ai+bj+ck.
i thick here is the catch , a logic problem.
 
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Pete:
sorry i wasnot able to say it clear. what i wanted to say is that
in effect you cannot choose the 2 angles of the 3. arbitrarily
for example
if two of the 3 angbles are both too small, the two cosine values' sum will be larger than 1, thus, you cannot get the last one,either.
 
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Pete:
sorry i wasnot able to say it clear. what i wanted to say is that
in effect you cannot choose the 2 angles of the 3. arbitrarily
for example
if two of the 3 angbles are both too small, the two cosine values' sum will be larger than 1, thus, you cannot get the last one,either.
Wow! I must still be pretty ill not to have seen that. :rofl: You're absolutely right of course. A good example is when the angle the unit vector makes with z is 90 degress. The two other degrees must then add up to 90 degrees. Thanks for pointing that out. I guess I'm getting old! :smile:

Pete
 
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zhanghe, Now I am clear, thanks. But this is the first time I can remember where we started with a wrong assumption (existence of a vector satistying the 4 conditions) and ended with a wrong answer (a neat answer for the vector even though the vector satisfying all the conditions may not exist). Generally, in math, when we start with a wrong assumption, we end with an absurd answer so we realise we started with a wrong assumption. (Reductio ad absurdum). Are there any examples of this kind i.e start with a wrong assumpiton and also end up with an incorrect or misleading answer.
 
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Pete and manjuvenamma ,you are welcome, :)
 
Simply resolve vector 'm' on any one axis n plane formed by other two...

lets say z-axis n xy-plane

This means,

on z-axis, we get " m.cosZ"

on xy-plane,we get "m.sinZ"

now further resolving oon xy-plane gives

m.sinZ.cosθ and m.sinZ.sinθ

and we have the following condition,

m.sinZ.cosθ = m.cosX
m.sinZ.sinθ = m.cosY

THIS GIVES

sinZ.cosθ = cosX
sinZ.sinθ = cosY


the relation u were searching for,..........

Please reply if anyone find it useful...
 

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