Linear Algebra - Use angles between vectors to find other vectors

In summary, the problem is to find all unit vectors in R^3 that make an angle of pi/4 radians with vector Y and an angle of pi/3 radians with vector Z. The solution involves setting up equations for the dot product of X with Y and Z, and solving for the components of X in terms of a variable x3. However, not all values of x3 will result in a unit vector, so further constraints need to be added to the equations.
  • #1
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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 = ([itex]x_{1}[/itex],[itex]x_{2}[/itex],[itex]x_{3}[/itex])

We need X*Y = cos([itex]\pi[/itex]/4) |x| |y|
and X*Z = cos([itex]\pi[/itex]/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= [itex]x_{1}[/itex] + [itex]x_{3}[/itex] = srt(2)/2 * sqrt(2) = 1
X*Z = [itex]x_{2}[/itex] = 1/2 * sqrt(2) = sqrt(2)/2

Solving for [itex]x_{1}[/itex] and [itex]x_{2}[/itex] in terms of [itex]x_{3}[/itex] we get:

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

Problem: X is not a unit vector for all [itex]x_{3}[/itex], 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 [itex]x_{1}[/itex]^2 + [itex]x_{2}[/itex]^2 + [itex]x_{3}[/itex]^2 = 1, but I'm not sure how to handle such a nonlinear constraint!
 
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  • #2
Heute said:

The Attempt at a Solution



let X = ([itex]x_{1}[/itex],[itex]x_{2}[/itex],[itex]x_{3}[/itex])

We need X*Y = cos([itex]\pi[/itex]/4) |x| |y|
and X*Z = cos([itex]\pi[/itex]/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= [itex]x_{1}[/itex] + [itex]x_{3}[/itex] = srt(2)/2 * sqrt(2) = 1
X*Z = [itex]x_{2}[/itex] = 1/2 * sqrt(2) = sqrt(2)/2
Why is there a factor of ##\sqrt{2}## in the equation for ##\vec{x}\cdot\vec{z}##?
 
  • #3
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.
 
  • #4
Apparently, I can't edit the original post, so I'll re-write it here:


The Attempt at a Solution



let X = ([itex]x_{1}[/itex],[itex]x_{2}[/itex],[itex]x_{3}[/itex])

We need X*Y = cos([itex]\pi[/itex]/4) |x| |y|
and X*Z = cos([itex]\pi[/itex]/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= [itex]x_{1}[/itex] + [itex]x_{3}[/itex] = srt(2)/2 * sqrt(2) = 1
X*Z = [itex]x_{2}[/itex] = 1/2 * 1 = 1/2

Solving for [itex]x_{1}[/itex] and [itex]x_{2}[/itex] in terms of [itex]x_{3}[/itex] we get:

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

Problem: X is not a unit vector for all [itex]x_{3}[/itex], 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 [itex]x_{1}[/itex]^2 + [itex]x_{2}[/itex]^2 + [itex]x_{3}[/itex]^2 = 1, but I'm not sure how to handle such a nonlinear constraint!
 
  • #5
You just need to find the value or values of x3 for which X has unit length. Solve the equation X2=1 for x3.
 

1. What is the purpose of using angles between vectors in linear algebra?

The use of angles between vectors in linear algebra allows us to determine the relationship between two vectors. It helps us understand the direction and magnitude of vectors and can be used to find other vectors through trigonometric calculations.

2. How do you calculate the angle between two vectors?

To calculate the angle between two vectors, we use the dot product formula: θ = cos^-1((A · B) / (|A| * |B|)), where A and B are the two vectors and |A| and |B| are their magnitudes. The resulting angle will be in radians.

3. Can you use angles between vectors to find linearly dependent or independent vectors?

Yes, you can use angles between vectors to determine if they are linearly dependent or independent. If the angle between two vectors is 0° or 180°, they are linearly dependent. If the angle is 90°, they are linearly independent. Any other angle indicates that the vectors are neither dependent nor independent.

4. How can angles between vectors be used to solve real-world problems?

Angles between vectors are often used in physics and engineering to solve problems involving forces and motion. For example, in projectile motion, the angle between the initial velocity vector and the acceleration vector can be used to calculate the trajectory of the object.

5. What is the relationship between angles and the unit circle in linear algebra?

The unit circle is a useful tool in linear algebra for understanding angles between vectors. The x and y coordinates on the unit circle can represent the cosine and sine of an angle, respectively. This allows us to use trigonometric functions to find angles between vectors and perform vector operations geometrically.

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