Linear Algebra - Use angles between vectors to find other vectors

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The discussion focuses on finding all unit vectors X in R^3 that form specific angles with given vectors Y and Z. The user sets up the equations using the dot product and the cosine of the angles, assuming the length of X is 1. However, they encounter difficulties in ensuring that X remains a unit vector while satisfying the angle conditions. The conversation highlights the need to solve for x3 to maintain unit length and suggests incorporating a nonlinear constraint to achieve this. Ultimately, the challenge lies in correctly determining the values of x3 that allow X to meet all specified requirements.
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Homework Statement



Find all the unit vectors X element of R^3 that make an angle of pi/4 radians with vector Y = (1,0,1) and an angle of pi/3 radians with vector Z = (0,1,0)


Homework Equations



For any two vectors X and Y element of R^n, the dot-prodict of X and Y is equals to the length of X times the length of Y times the cosine of the angle between X and Y. That is:

X*Y = cos(t)|x||y|

The Attempt at a Solution



let X = (x_{1},x_{2},x_{3})

We need X*Y = cos(\pi/4) |x| |y|
and X*Z = cos(\pi/3)|x||z|

We want the length of X, that is |x|, to be 1 so I'll assume that it is 1 for now (this could be a bad idea).

X*Y= x_{1} + x_{3} = srt(2)/2 * sqrt(2) = 1
X*Z = x_{2} = 1/2 * sqrt(2) = sqrt(2)/2

Solving for x_{1} and x_{2} in terms of x_{3} we get:

X = (1, 1/2, 0) + x_{3}(-1, 0, 1)

Problem: X is not a unit vector for all x_{3}, so we haven't really found a formula for "all unit vectors" which fulfill the initial requirements. This is where I'm stuck! I thought about including another variable for |x| in the restrictions and adding a further restriction that x_{1}^2 + x_{2}^2 + x_{3}^2 = 1, but I'm not sure how to handle such a nonlinear constraint!
 
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Heute said:

The Attempt at a Solution



let X = (x_{1},x_{2},x_{3})

We need X*Y = cos(\pi/4) |x| |y|
and X*Z = cos(\pi/3)|x||z|

We want the length of X, that is |x|, to be 1 so I'll assume that it is 1 for now (this could be a bad idea).

X*Y= x_{1} + x_{3} = srt(2)/2 * sqrt(2) = 1
X*Z = x_{2} = 1/2 * sqrt(2) = sqrt(2)/2
Why is there a factor of ##\sqrt{2}## in the equation for ##\vec{x}\cdot\vec{z}##?
 
Apologies! That sqrt(2) should be a 1 (the length of Z). I copied that sqrt(2) from the memory of another similar problem I was working on. I'll edit the original post!

This doesn't, however fundamentally change things. The question still remains.
 
Apparently, I can't edit the original post, so I'll re-write it here:


The Attempt at a Solution



let X = (x_{1},x_{2},x_{3})

We need X*Y = cos(\pi/4) |x| |y|
and X*Z = cos(\pi/3)|x||z|

We want the length of X, that is |x|, to be 1 so I'll assume that it is 1 for now (this could be a bad idea).

X*Y= x_{1} + x_{3} = srt(2)/2 * sqrt(2) = 1
X*Z = x_{2} = 1/2 * 1 = 1/2

Solving for x_{1} and x_{2} in terms of x_{3} we get:

X = (1, 1/2, 0) + x_{3}(-1, 0, 1)

Problem: X is not a unit vector for all x_{3}, so we haven't really found a formula for "all unit vectors" which fulfill the initial requirements. This is where I'm stuck! I thought about including another variable for |x| in the restrictions and adding a further restriction that x_{1}^2 + x_{2}^2 + x_{3}^2 = 1, but I'm not sure how to handle such a nonlinear constraint!
 
You just need to find the value or values of x3 for which X has unit length. Solve the equation X2=1 for x3.
 
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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