Rotating a Vector in a Coordinate System: Explaining the Formula

In summary, the conversation discusses how to find the new components of a vector after rotating the coordinate axes by an angle theta. Three different methods are presented: using trigonometry, using vector algebra, and using a distance formula. Each method involves finding the components of the original vector in the new coordinate system and using trigonometric identities to determine the new components. The conversation also includes a discussion on the derivation of the distance formula and a clarification on the subscript in one of the equations.
  • #1
misogynisticfeminist
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Say, i have a vector a, defined in a coordinate system, x-y-z and i rotate the axes by an angle theta around the z-axis, so i have my z-component invariant in this change of basis. Can someone show me why,

[tex] a_x' = a_x cos\theta + a_y sin\theta [/tex] and

[tex] a_x' = -a_x sin\theta + a_y cos\theta [/tex]

i can't seem to find any place where they show how this is done.
 
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  • #2
Draw a vector and two sets of orthogonal axes.
For each set of axes, form a right triangles with the vector as the hypotenuse, and the legs parallel to the axes.
Do a little trigononometry.
I'll post a URL with a picture, if I can find one.

Ok here's one:
http://web.umr.edu/~oci/Topic12/T12-5/cml12-5a/frame.html [Broken]
(although the vector is not drawn from the origin to the little box).

(Check the subscript on the LHS of your second equation.)
 
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  • #3
The new x-axis is a line tilted at some angle, [itex]\theta[/tex], to the horizontal. Basically, if you want [itex]a'_y[/tex], you want the distance between a point and this line. The formula for this is
[tex]|a'_y|= \frac {|ma_x+b-a_y|} {\sqrt{m^2+1}}[/tex]
where the equation of the line is [itex]y=mx+b[/tex]. Since the line passes through the origin, b=0. [itex]tan(\theta)=m[/tex], so:
[tex]|a'_y|= \frac {|tan(\theta)a_x-a_y|} {\sqrt{tan(\theta)^2+1}}[/tex]
[tex]= \frac {|tan(\theta)a_x-a_y|} {|sec(\theta)|}[/tex]

[tex]= |tan(\theta)a_x-a_y|\times|cos(\theta)|[/tex]

[tex]=|sin(\theta)a_x-cos(\theta)a_y|[/tex]

Since, for [itex]\theta=0[/tex], we should have [itex]a'_y=a_y[/tex], this means the signs will only work out properly if:
[tex]a'_y=-sin(\theta)a_x+cos(\theta)a_y[/tex]
The other equation is found similarly. Let me know if you want to see the derivation of this distance formula.
 
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  • #4
Here's a vector-algebraic proof.

For the first set of axes, we have unit vectors [tex]\hat x[/tex] and [tex]\hat y[/tex].
For the second set, which is rotated counterclockwise by angle [tex]\theta[/tex], we have [tex]\hat x'[/tex] and [tex]\hat y'[/tex].

Note that [tex]A_x= \vec A\cdot \hat x[/tex] and so on.

Now, compute [tex]A_{x'}= \vec A\cdot \hat x'[/tex] where one writes
[tex] \vec A=(\vec A \cdot \hat x) \hat x + (\vec A \cdot \hat y) \hat y=A_x\hat x + A_y\hat y[/tex].
So, let's plug it in:
[tex]\begin{align*}
A_{x'}
&= \vec A\cdot \hat x' \\
&= \left( A_x\hat x + A_y\hat y \right) \cdot \hat x' \\
&=(A_x) \hat x\cdot \hat x' + (A_y) \hat y\cdot \hat x' \\
&=(A_x) \cos\theta + (A_y) \cos(90^\circ - \theta) \\
&=(A_x) \cos\theta + (A_y) \sin\theta \\
\end{align}
[/tex]
Do the same for [tex]A_{y'}[/tex], and note that [tex] \hat x\cdot \hat y' =\cos(90^\circ+\theta)= -\sin\theta [/tex] and [tex] \hat y\cdot \hat y' =\cos\theta[/tex].
 
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  • #5
The original vector had an x-component and a y-component. I find it easiest to just follow those two components separately.

When the original vector gets rotated the original x-component, let’s call that X, will not be only in the direction of the x-axis anymore, the component of the rotated “original x-component“ that will still be in the direction of the x-axis is:
[tex]X\cos(\theta)[/tex]
but it will now also have a component in the direction of the y-axis, which will be:
[tex]-X\sin(\theta)[/tex]

The same happens with the original y component. The component of the rotated “original y component“ that will still be in the direction of the y-axis is:
[tex]Y\cos(\theta)[/tex]
and the component in the direction of the x-axis will be:
[tex]Y\sin(\theta)[/tex]

To get the x-component of the new vector add the parts in the direction of the x-axis:
[tex]X\cos(\theta) + Y\sin(\theta)[/tex]
To get the y-component of the new vector add the parts in the direction of the y-axis:
[tex]-X\sin(\theta) + Y\cos(\theta)[/tex]
 
  • #6
hey thanks a lot for the help. Yup the subscript on the LHS should be y. Here's what i have done, a little algebra,

[tex] a_x' cos \theta = a_x[/tex]

[tex] a_x' sin \theta= a_y [/tex]

what i did was to multiply sin and cosine to both equations respectively and add them up, to get the expression for [tex] a_x'[/tex]

for [tex] a_y' [/tex] i did the same thing, except that this time, comlementary angles were involved and the vector a_x points in the opposite direction.

Is that right?

edit: i'll look at the rest of the methods too. : )
 
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1. What is a vector and why is it important in a coordinate system?

A vector is a mathematical object that represents a quantity with both magnitude and direction. In a coordinate system, vectors are used to describe the position and movement of objects. They are important because they allow us to perform calculations and analyze data in a geometric way.

2. How do you rotate a vector in a coordinate system?

To rotate a vector in a coordinate system, you can use a formula that involves trigonometric functions. The formula is: x' = x * cos(theta) - y * sin(theta) and y' = x * sin(theta) + y * cos(theta), where (x,y) are the original coordinates of the vector, (x',y') are the coordinates after rotation, and theta is the angle of rotation.

3. Why do we use the formula to rotate a vector instead of simply changing the angle?

The formula for rotating a vector allows us to easily calculate the new coordinates of the vector without having to manually change the angle. It also works for any angle of rotation, not just specific ones.

4. Can you explain the concept of a unit vector and its role in rotating a vector?

A unit vector is a vector with a magnitude of 1 and is often used to represent direction. In rotating a vector, we can use unit vectors to determine the direction of the rotation. For example, if we want to rotate a vector counterclockwise, we can use a unit vector in the positive direction.

5. Are there any real-world applications of rotating a vector in a coordinate system?

Yes, there are many real-world applications of rotating a vector in a coordinate system. For example, in computer graphics, vectors are used to represent objects and rotating them allows for the creation of 3D animations. In physics and engineering, rotating vectors are used to study the movement of objects and analyze forces. They are also used in navigation systems to determine the direction and position of objects.

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