How can I determine an orthogonal vector to a given vector in 3D space?

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To find an orthogonal vector in 3D space, one can use the cross product between the given directional vector and a standard up vector, such as <0, 1, 0>. This method generates a horizontal vector, which can then be crossed with the original directional vector to derive the proper up vector. The discussion highlights the importance of ensuring that the chosen up vector is not parallel to the directional vector to maintain orthogonality. The user expresses a need for 3D plotting software to verify the results of their calculations. Overall, the approach of using cross products is confirmed as a valid method for determining orthogonal vectors.
rocketman123
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Hey guys,

Given a vector, ie < -1, 2, 3 > , how does one go about finding a vector which is orthogonal to it?

I also have another vector < x, y ,z > which is the point of origin for the above vector.

In context, I'm given a directional vector from which I need to find an 'up' vector and a 'horizontal' vector. You can see here http://www.cs.auckland.ac.nz/~jli023/images/opengl/pov-ray/viewplaneAnglechanged.jpg - I have a 'look_at' vector and must determine a suitable up and right vector.

I know that to get the right/ horiztonal vector I can just take the cross product between the directional / look at vector and the up vector. However, how to get the up vector confuses me.

A standard up vector is <0 1 0 >. Would it make sense to take the cross product of <0 1 0 > and the direction vector - to get the horizontal vector. And then take the cross product of the horizontal and directional vectors to get the proper up vector? It makes sense to me, however I have no real way of checking if my answer is correct! - I need to find some nice 3d plotting software hehe

Cheers,
Dan
 
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Welcome to PF!

Hi Dan! Welcome to PF! :smile:

I'm confused :confused: … surely all up vectors are the same?

(and not orthogonal to the 'look at' vector)
 
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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