# Transformation of Vectors in a Rotated Coordinate System

• domabo
In summary, the conversation discusses a given Cartesian coordinate system and a vector within it. A new coordinate system is introduced that is rotated by 60 degrees with respect to the original system. The components of the vector in the new system are calculated using equations (3.3.20) and (3.3.21). The magnitude of the vector is also calculated using both sets of components, and it is found to be the same. However, there is a mistake in the derivation of the components in the new system. The correct expressions can be found using equations (3.3.22) and (3.3.23). The conversation ends with a question about the geometric property that allows for the switch in sine
domabo

## Homework Statement

With respect to a given Cartesian coordinate system S , a vector A has components Ax= 5 , Ay= −3 , Az = 0 . Consider a second coordinate system S′ such that the (x′, y′) x y z coordinate axes in S′ are rotated by an angle θ = 60 degrees with respect to the (x, y) coordinate axes in S , (Figure 3.26). (a) What are the components Ax and Ay of vector A in coordinate system S′ ? (b) Calculate the magnitude of the vector using the ( Ax , A y ) components and using the ( Ax' , Ay' ) components. Does your result agree with what you expect?
https://imgur.com/gallery/b06Nn

## Homework Equations

i' = icosθ + jsinθ (3.3.20)
j' = -isinθ + jcosθ (3.3.21)
(x,y) -> (x',y')
and
r = xi + yj -> r = x'i' + y'j'
where
x' = xcosθ + ysinθ ,
y’ = xsinθ - ycosθ ,
i' = icosθ + jsinθ ,
and
j' = -isinθ + jcosθ

## The Attempt at a Solution

I can see how the problem works, but, if you click on the link to see the attached photos, I'm having more trouble with their derivations. I believe that I can see how the i' and j' were derived. I attached what I attempted as a photo at the end of the gallery. I do not understand how, in their derivation, they arrived at
x' = xcosθ + ysinθ and y’ = xsinθ - ycosθ. More specifically, it is the negative ycos that troubles me. The way I saw it was using the prior derivations and then finally grouping the terms that matched with i' and j', you would get y' = -xsinθ + ycosθ.In the given problem, the Aycos is negative, which makes more sense to me. I really would like to understand the underpinnings of this because it's crucial moving forward. Thank you so much.

If you can see why the following two equations are true, the rest is not very complicated.
##\hat{i}'=\hat{i}\cos \theta+\hat{j}\sin\theta~~~## (3.3.20)
##\hat{j}'=-\hat{i}\sin \theta+\hat{j}\cos\theta~~~## (3.3.21)
Now a vector in the primed coordinate system is written as
##\vec{A}=A'_x \hat{i}'+A'_y \hat{j}'##
Replace the primed unit vectors to get
##\vec{A}=A'_x (\hat{i}\cos \theta+\hat{j}\sin\theta)+A'_y (-\hat{i}\sin \theta+\hat{j}\cos\theta)##
Gather like terms together to get
##\vec{A}= (A'_x\cos \theta-A'_y\sin \theta)\hat{i}+(A'_x\sin\theta+A'_y\cos\theta)\hat{j}##
Clearly, the components of ##\vec{A}## in the unprimed frame are
##A_x=(A'_x\cos \theta-A'_y\sin \theta)##
##A_y=(A'_x\sin\theta+A'_y\cos\theta)##
You can go backwards by starting with (3.3.22) and (3.3.23) and by following the same procedure. Also note that in the special case ##\vec{A} = \vec{r}##, you have ##A'_x=x'##, ##A'_y=y'##, ##A_x=x##, ##A_y=y##.

domabo said:
I can see how the problem works, but, if you click on the link to see the attached photos, I'm having more trouble with their derivations. I believe that I can see how the i' and j' were derived. I attached what I attempted as a photo at the end of the gallery. I do not understand how, in their derivation, they arrived at
x' = xcosθ + ysinθ and y’ = xsinθ - ycosθ. More specifically, it is the negative ycos that troubles me. The way I saw it was using the prior derivations and then finally grouping the terms that matched with i' and j', you would get y' = -xsinθ + ycosθ.
They made a mistake in going from the first line of (3.3.24) to the second line of (3.3.24). So, they get the wrong result in (3.3.26). The signs of the two terms on the right side of (3.3.26) are wrong. Note that they follow this derivation with a different derivation where the correct expression is arrived at in (3.3.32).

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kuruman said:
If you can see why the following two equations are true, the rest is not very complicated.
##\hat{i}'=\hat{i}\cos \theta+\hat{j}\sin\theta~~~## (3.3.20)
##\hat{j}'=-\hat{i}\sin \theta+\hat{j}\cos\theta~~~## (3.3.21)
Now a vector in the primed coordinate system is written as
##\vec{A}=A'_x \hat{i}'+A'_y \hat{j}'##
Replace the primed unit vectors to get
##\vec{A}=A'_x (\hat{i}\cos \theta+\hat{j}\sin\theta)+A'_y (-\hat{i}\sin \theta+\hat{j}\cos\theta)##
Gather like terms together to get
##\vec{A}= (A'_x\cos \theta-A'_y\sin \theta)\hat{i}+(A'_x\sin\theta+A'_y\cos\theta)\hat{j}##
Clearly, the components of ##\vec{A}## in the unprimed frame are
##A_x=(A'_x\cos \theta-A'_y\sin \theta)##
##A_y=(A'_x\sin\theta+A'_y\cos\theta)##
You can go backwards by starting with (3.3.22) and (3.3.23) and by following the same procedure. Also note that in the special case ##\vec{A} = \vec{r}##, you have ##A'_x=x'##, ##A'_y=y'##, ##A_x=x##, ##A_y=y##.
Thank you for your thorough response. I would like to double check if I understand if (3.3.20) and (3.3.21) are true. Is it because they're showing all components in the respective ##\hat{i}'## and ##\hat{j}'## directions? By what geometry law/property does that one angle rule where sine becomes representative of the x-component and cosine becomes representative of the y-component? Sorry if I'm not entirely clear or this isn't making sense... I do understand sine and cosine ultimately represent the ratios of SOH and CAH but what geometric property allows for some new angle to be the same as the original and then allows for this switch in sine and cosine? (This also appears with blocks on inclines if that clears up what I'm trying to say.) Thanks.

Also, why doesn't our final showing of the vector in the primed coordinate system look like ##\vec{A}=A'_x \hat{i}'+A'_y \hat{j}'## but instead are written in terms of ##A_x## and ##A_y## ?
My apologies if this is a lot.

domabo said:
By what geometry law/property does that one angle rule where sine becomes representative of the x-component and cosine becomes representative of the y-component?
I am not sure exactly what you mean, but here is a pictorial approach. Study the figure I have attached. Then answer the following questions. If something confuses you, let me know. The picture shows two coordinate systems. The red axes are rotated by angle θ relative to the black axes. Ignore the blue vector A for the moment.
1. Do you see why the angles that I have labeled are all θ, the same as the angle between unit vectors i and i'?
2. Do you see why the red hypotenuses of the right triangles have magnitudes equal to 1?
3. Do you see why the right sides of these triangles (dotted lines) are labeled by trig functions as they are?
4. Do you see why the x-component of i' is cosθ and its y-component is sinθ?
5. Do you see why ##\hat{i}'=\hat{i}\cos \theta+\hat{j}\sin\theta~##? This is (3.3.20).
6. Do you see why the x-component of j' is -sinθ and its y-component is cosθ?
7. Do you see why ##\hat{j}'=-\hat{i}\sin \theta+\hat{j}\cos\theta~##? This is (3.3.21).

Now take a good look at the blue vector A. If you draw the perpendicular dotted lines from its tip, you will see that it has one set of components in the black frame and another set of components in the red frame. If you know the values of the components in the red frame, namely A'x and A'y, then the components in the black frame, namely Ax and Ay, are obtained from the equations I gave you
##A_x=(A'_x\cos \theta-A'_y\sin \theta)##
##A_y=(A'_x\sin\theta+A'_y\cos\theta)##

The key idea is that a vector can only be expressed if there is an agreed upon coordinate system in which one writes its components. Different coordinate systems provide different components for the same vector.

#### Attachments

• Axes.png
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kuruman said:
I am not sure exactly what you mean, but here is a pictorial approach. Study the figure I have attached. Then answer the following questions. If something confuses you, let me know. The picture shows two coordinate systems. The red axes are rotated by angle θ relative to the black axes. Ignore the blue vector A for the moment.
1. Do you see why the angles that I have labeled are all θ, the same as the angle between unit vectors i and i'?
2. Do you see why the red hypotenuses of the right triangles have magnitudes equal to 1?
3. Do you see why the right sides of these triangles (dotted lines) are labeled by trig functions as they are?
4. Do you see why the x-component of i' is cosθ and its y-component is sinθ?
5. Do you see why ##\hat{i}'=\hat{i}\cos \theta+\hat{j}\sin\theta~##? This is (3.3.20).
6. Do you see why the x-component of j' is -sinθ and its y-component is cosθ?
7. Do you see why ##\hat{j}'=-\hat{i}\sin \theta+\hat{j}\cos\theta~##? This is (3.3.21).

Now take a good look at the blue vector A. If you draw the perpendicular dotted lines from its tip, you will see that it has one set of components in the black frame and another set of components in the red frame. If you know the values of the components in the red frame, namely A'x and A'y, then the components in the black frame, namely Ax and Ay, are obtained from the equations I gave you
##A_x=(A'_x\cos \theta-A'_y\sin \theta)##
##A_y=(A'_x\sin\theta+A'_y\cos\theta)##

The key idea is that a vector can only be expressed if there is an agreed upon coordinate system in which one writes its components. Different coordinate systems provide different components for the same vector.
View attachment 218194
Thank you so much. I went through the entire process again and came up with the same result as you. Just to be clear on a final note, saying ##A_x## is the same as saying those components in the ##\hat{i}## direction and saying ##A_y## is the same as saying ##\hat{j}##? Also, why is it that this general result works with the example problem if the Vector A is along one of the new axes? Does this general derived result work for all angle transformations? I very much appreciate the image you provided and kindly ask if you may do the same for this example problem because when I try to drop a perpendicular to show the components, I'm left missing one piece and only have the vector A and an x component. I can see the algebraic manipulation for obtaining the results for the example problem but don't see the proper picture for it. I apologize if I may be asking a lot. I realize a lot of physics is discovering on your own but I've been struggling for a while with this and need some guidance; hence my asking of these questions.

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Study the drawing. It shows how to draw the components of A in the primed (red) and unprimed (black) frames.

#### Attachments

• Axes_2.png
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kuruman said:
Study the drawing. It shows how to draw the components of A in the primed (red) and unprimed (black) frames.
View attachment 218240
Yes, I see this. My question, however, is how would I represent this for the example problem where the vector is along one of the (rotated) axes? Where do I drop the line perpendicular to the rotated axes for that situation? Also, thank you for your continued patience and replies; I appreciate it very much.

When a vector is along one of the axes, then its magnitude is the component along that axis while the other component is zero. When you can't draw the perpendicular, then it is zero and you don't draw it.

In the black (unprimed) frame,
##\vec{A}=0\hat{i}+A\hat{j}##, that is ##A_x = 0##, ##A_y = A##.

In the red (primed) frame,
##\vec{A}=A\sin \theta\hat{i}+A \cos \theta\hat{j}##, that is ##A'_x = A\sin \theta##, ##A'_y = A \cos \theta##.

#### Attachments

• Axes_3.png
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kuruman said:
When a vector is along one of the axes, then its magnitude is the component along that axis while the other component is zero. When you can't draw the perpendicular, then it is zero and you don't draw it.
View attachment 218242

In the black (unprimed) frame,
##\vec{A}=0\hat{i}+A\hat{j}##, that is ##A_x = 0##, ##A_y = A##.

In the red (primed) frame,
##\vec{A}=A\sin \theta\hat{i}+A \cos \theta\hat{j}##, that is ##A'_x = A\sin \theta##, ##A'_y = A \cos \theta##.
Then how is it that they came to the same conclusion as we discussed before with this new vector, which clearly lies along the j' axis? https://imgur.com/gf7dVgm

domabo said:
... which clearly lies along the j' axis?
It's probably badly drawn and not to scale. Look at figure 3.25. Vector r is not along any of the primed or unprimed axes. Trust the math and not the figures.

On edit: Actually, figure 3.26 is drawn to scale and I take back what I said in this post. If you calculate a number for the primed x-component as shown , you get A'x = -0.1, which is pretty close to zero. I can see the source of your confusion (and mine).

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kuruman said:
It's probably badly drawn and not to scale. Look at figure 3.25. Vector r is not along any of the primed or unprimed axes. Trust the math and not the figures.

On edit: Actually, figure 3.26 is drawn to scale and I take back what I said in this post. If you calculate a number for the primed x-component as shown , you get A'x = -0.1, which is pretty close to zero. I can see the source of your confusion (and mine).
I thought I had it but I can't settle for not being 100% certain. How would I draw a perpendicular for Figure 3.26? Sorry if this is too much.

domabo said:
I thought I had it but I can't settle for not being 100% certain. How would I draw a perpendicular for Figure 3.26? Sorry if this is too much.
I will let you draw the perpendicular. Here a scaled drawing of Figure 3.26. Note that the negative y' axis (red dotted line) forms an angle of 1° with vector A. Do you understand the problem now?

On edit: I had to replace the figure because I misidentified the axes relative to Figure 3.26.

#### Attachments

• Axes_4.png
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kuruman said:
I will let you draw the perpendicular. Here a scaled drawing of Figure 3.26. Note that the negative y' axis (red dotted line) forms an angle of 1° with vector A. Do you understand the problem now?

On edit: I had to replace the figure because I misidentified the axes relative to Figure 3.26.
View attachment 218264
Where does it say anything about 1 degree? Or did you just draw that so I would see that the vector lies along the negative y' axis? Here's an image of what I drew. https://imgur.com/VZgzZER

domabo said:
Where does it say anything about 1 degree?
It is not explicitly stated but it is implicit in the givens of the question.
(a) According to Figure 3.26, the dotted negative y'-axis is 30° below the horizontal x-axis because its perpendicular x'-axis is at 60° to the left of the vertical y-axis.
(b) It is given that Ax = 5 and Ay = -3. The angle that vector A makes relative to the horizontal x-axis is given by ##\varphi = \arctan(-3/5) = -31^o##. The minus sign means "below the positive x-axis". Thus, the difference between the two is 1°.

kuruman said:
It is not explicitly stated but it is implicit in the givens of the question.
(a) According to Figure 3.26, the dotted negative y'-axis is 30° below the horizontal x-axis because its perpendicular x'-axis is at 60° to the left of the vertical y-axis.
(b) It is given that Ax = 5 and Ay = -3. The angle that vector A makes relative to the horizontal x-axis is given by ##\varphi = \arctan(-3/5) = -31^o##. The minus sign means "below the positive x-axis". Thus, the difference between the two is 1°.
And my picture then fails to accurately represent the situation?

domabo said:
And my picture then fails to accurately represent the situation?
Your picture doesn't show the components of A in the primed frame.

kuruman said:
Your picture doesn't show the components of A in the primed frame.
Are they there and I've failed to label them as so? Also, why was the one degree thing even relevant if the solution did not mention that at all?

domabo said:
Are they there and I've failed to label them as so?
They are not there. The perpendiculars used to find the components of A must start at the tip of A and end at one of the appropriate axis. Your figure shows no perpendiculars at the tip of A. I showed you how this is done in post #7 and you replied
domabo said:
Yes, I see this.
Did you really "see this"?
domabo said:
Also, why was the one degree thing even relevant if the solution did not mention that at all?
The 1° angle is relevant because it has to do with your repeated assertion that vector A is along one of the axes. That assertion is unfounded. I provided the reason why. It looks like A is along the negative y'-axis but it is not.

kuruman said:
They are not there. The perpendiculars used to find the components of A must start at the tip of A and end at one of the appropriate axis. Your figure shows no perpendiculars at the tip of A. I showed you how this is done in post #7 and you replied
Did you really "see this"?

The 1° angle is relevant because it has to do with your repeated assertion that vector A is along one of the axes. That assertion is unfounded. I provided the reason why. It looks like A is along the negative y'-axis but it is not.
If the vector weren't along A, and is one degree off as you say, why wouldn't their diagram reflect that by showing the negative portion of the j' axis? I can see now, with the one degree, how the perpendicular is formed. Yes, I did really see it; I just can't see how we can apply the results of the first derivation to this problem. It doesn't make sense to me how that one degree wasn't reflected in their solution of the problem at all. They used the angle 60 degrees and never mentioned one.

domabo said:
If the vector weren't along A ...
I don't understand. A is the vector.
domabo said:
It doesn't make sense to me how that one degree wasn't reflected in their solution of the problem at all. They used the angle 60 degrees and never mentioned one.
The 60° is the angle by which the primed coordinate axes are rotated relative to the unprimed axes. That's all that is needed to mention. Look at equations (3.3.35) and (3.3.36). They allow you to find the components of A in the primed frame if all you know is the components of A in the unprimed frame and the angle by which one frame is rotated relative to the other. You don't need to know any angle that A forms with the primed axes, i.e. the 1°. That's the whole point of this exercise and that's why they did not mention it. I mentioned it for the reasons that I already explained in post #19.

kuruman said:
I don't understand. A is the vector.
The 60° is the angle by which the primed coordinate axes are rotated relative to the unprimed axes. That's all that is needed to mention. Look at equations (3.3.35) and (3.3.36). They allow you to find the components of A in the primed frame if all you know is the components of A in the unprimed frame and the angle by which one frame is rotated relative to the other. You don't need to know any angle that A forms with the primed axes, i.e. the 1°. That's the whole point of this exercise and that's why they did not mention it. I mentioned it for the reasons that I already explained in post #19.

Ok. I meant if the vector A weren’t along j’ but I’m sure you knew that.

My whole point in dragging this out is: I want to be able to find the components of A in the primed coordinate system just using the picture and nothing else.

domabo said:
I want to be able to find the components of A in the primed coordinate system just using the picture and nothing else.
Why didn't you say so? To do what you want, you need a good picture and a good strategy.
Here's a similar problem in which the vector is not close to any axis so that there will be no confusion, but the coordinate frames are the same..
Given that ##\vec{B}=4 \hat{i}+2\hat{j}##, find ##B_{x'}## and ##B_{y'}##. The figure below is drawn to scale.

Solution without using the formulas developed in the exercise
First find the angle ##\alpha## that ##\vec{B}## forms with respect to the x-axis.
##\alpha=\arctan(2/4)=26.5^o##
Now find the angle ##\beta## that the vector forms with respect to the x'-axis.
Since the angle between the x-axis and the x'-axis is given as 60o,
##\alpha+\beta=60^o~~\rightarrow~~\beta=60^o-\alpha=60^o-26.5^o=33.5^o.##
The angle between the positive x'-axis and ##\vec{B}## in a counterclockwise (positive) direction is ##\gamma =360^o-\beta= 326.5^o.## (Angle ##\gamma## is not shown to avoid clutter in the drawing.)
Now find the components of ##\vec{B}## in the primed frame the usual way.
##\vec{B}=B\cos\gamma~ \hat{i}'+B\sin\gamma~\hat{j}'##
With ##B=\sqrt{4^2+2^2}=4.47##, ##\cos(326.5^o)=0.834## and ##\sin(326.5^o)=-0.552##,
##\vec{B}=4.47\times0.834~ \hat{i}'+4.47\times (-0.552)~\hat{j}'=3.73\hat{i}'-2.47~\hat{j}'##

Solution using the formulas developed in the exercise
##B_{x'}=B_x\cos\theta+B_y\sin\theta=4\times \cos60^o+2\times\sin60^o=2+1.73=3.73.##
##B_{y'}=-B_x\sin\theta+B_y\cos\theta=-4\times \sin60^o+2\times\cos60^o=-3.46+1.00=-2.46.##

The two methods agree to within round off errors. Which one do you think is easier to implement?

#### Attachments

• Axes_5.png
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kuruman said:
Why didn't you say so? To do what you want, you need a good picture and a good strategy.
Here's a similar problem in which the vector is not close to any axis so that there will be no confusion, but the coordinate frames are the same..
Given that ##\vec{B}=4 \hat{i}+2\hat{j}##, find ##B_{x'}## and ##B_{y'}##. The figure below is drawn to scale.
View attachment 218289
Solution without using the formulas developed in the exercise
First find the angle ##\alpha## that ##\vec{B}## forms with respect to the x-axis.
##\alpha=\arctan(2/4)=26.5^o##
Now find the angle ##\beta## that the vector forms with respect to the x'-axis.
Since the angle between the x-axis and the x'-axis is given as 60o,
##\alpha+\beta=60^o~~\rightarrow~~\beta=60^o-\alpha=60^o-26.5^o=33.5^o.##
The angle between the positive x'-axis and ##\vec{B}## in a counterclockwise (positive) direction is ##\gamma =360^o-\beta= 326.5^o.## (Angle ##\gamma## is not shown to avoid clutter in the drawing.)
Now find the components of ##\vec{B}## in the primed frame the usual way.
##\vec{B}=B\cos\gamma~ \hat{i}'+B\sin\gamma~\hat{j}'##
With ##B=\sqrt{4^2+2^2}=4.47##, ##\cos(326.5^o)=0.834## and ##\sin(326.5^o)=-0.552##,
##\vec{B}=4.47\times0.834~ \hat{i}'+4.47\times (-0.552)~\hat{j}'=3.73\hat{i}'-2.47~\hat{j}'##

Solution using the formulas developed in the exercise
##B_{x'}=B_x\cos\theta+B_y\sin\theta=4\times \cos60^o+2\times\sin60^o=2+1.73=3.73.##
##B_{y'}=-B_x\sin\theta+B_y\cos\theta=-4\times \sin60^o+2\times\cos60^o=-3.46+1.00=-2.46.##

The two methods agree to within round off errors. Which one do you think is easier to implement?
I believe that I am starting to see what you mean, given your newest example.

A few questions remain. All have to do with what's in the attached image here: https://imgur.com/a/Z1wMA
Namely,
1) Is the drawing now correct? I have made another attempt at the example problem and used the method you just described (and have described earlier)... that is, using the inverse tangent to find the angle, then using the magnitude to find the primed components of A, and they do seem to match up with the solution.
2) Why is it that, in their solution, the final notation is ##\vec{A} = A'_x \hat{i} + A'_y\hat{j}## instead of ##\vec{A} = A'_x \hat{i}' + A'_y\hat{j}'## ? In other words, why don't we have the primes together? If I recall correctly, that is how they described the notation earlier... that the primes must be with one another?
3) I still don't understand how in their derivation they obtained the -ycos... I attached a screenshot of what I'm referencing in the newest image above but it's also in the original album of images.
Thank you.

domabo said:
1) Is the drawing now correct?
The drawing is now correct (but not to scale).
domabo said:
2) Why is it that, in their solution, the final notation is##\vec{A} = A'_x \hat{i} + A'_y\hat{j}## instead of ⃗##\vec{A} = A'_x \hat{i}' + A'_y\hat{j}'## ?
That's because whoever wrote this up forgot to put primes on the unit vectors. It's a typo.
domabo said:
3) I still don't understand how in their derivation they obtained the -ycos... I attached a screenshot of what I'm referencing in the newest image above but it's also in the original album of images.
It's another typo. See post #3 by @TSny.

kuruman said:
The drawing is now correct (but not to scale).

That's because whoever wrote this up forgot to put primes on the unit vectors. It's a typo.

It's another typo. See post #3 by @TSny.

Oh, ok. Thank you so much for your patience. I’ve definitely learned a lot.

## 1. What is meant by the transformation of vectors in a rotated coordinate system?

The transformation of vectors in a rotated coordinate system refers to the process of changing the representation of a vector from one coordinate system to another that is rotated with respect to the original system. This is necessary when analyzing vectors in three-dimensional space, as different coordinate systems may be used to describe the same vector.

## 2. Why is it important to understand the transformation of vectors in a rotated coordinate system?

Understanding the transformation of vectors in a rotated coordinate system is important in many fields of science and engineering, such as physics, mechanics, and computer graphics. It allows for accurate analysis and prediction of the behavior of objects in three-dimensional space, and is essential for solving complex problems involving rotations and transformations.

## 3. How is a vector transformed in a rotated coordinate system?

A vector is transformed in a rotated coordinate system using a transformation matrix, which is a mathematical tool that represents the relationship between the original coordinate system and the rotated coordinate system. The vector's components are multiplied by this matrix to obtain its new representation in the rotated system.

## 4. What is the difference between active and passive transformation of vectors in a rotated coordinate system?

In active transformation, the coordinate system is rotated while the vector remains fixed. This means that the vector's components are transformed to match the new coordinate system, but the physical orientation of the vector in space remains the same. In passive transformation, the vector is rotated while the coordinate system remains fixed, resulting in a change in both the vector's components and its orientation in space.

## 5. Can the transformation of vectors in a rotated coordinate system be applied to non-linear transformations?

Yes, the transformation of vectors in a rotated coordinate system can be applied to non-linear transformations, as long as the transformation can be expressed as a mathematical function. This includes rotations, translations, scaling, and shearing. However, more complex non-linear transformations may require more advanced mathematical tools, such as quaternions, to accurately represent the transformation.

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