ISRO2011-76

Question

What is the matrix that represents rotation of an object by $\theta^0$ about the origin in $\text{2D}?$

• A.

  $\cos \theta$	$- \sin \theta$
  $\sin \theta$	$\cos \theta$

• B.

  $\sin \theta$	$- \cos \theta$
  $\cos \theta$	$\sin \theta$

• C.

  $\cos \theta$	$- \sin \theta$
  $\cos \theta$	$\sin \theta$

• D.

  $\sin \theta$	$- \cos \theta$
  $\cos \theta$	$\sin \theta$

Comments

both $\begin{bmatrix} cos\theta &sin\theta \\ -sin\theta & cos\theta \end{bmatrix}$

and $\begin{bmatrix} cos\theta &-sin\theta \\ sin\theta & cos\theta \end{bmatrix}$

are possible bcz rotation may be clock wise and anti clock wise.

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Yes sir

First one is anticlockwise and other one is clockwise

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is it a gate question?

Answer 1

In rotation, we rotate the object at particular angle θ (theta) from its origin. From the following figure, we can see that the point P(X, Y) is located at angle φ from the horizontal X coordinate with distance r from the origin.

Let us suppose you want to rotate it at the angle θ. After rotating it to a new location, you will get a new point P’ (X’, Y’).

Using standard trigonometric the original coordinate of point P(X, Y) can be represented as −

$X= r\ cos\phi$           ....(1)

$Y= r\ sin\phi$        ........(2)

Same way we can represent the point P’ (X’, Y’) as −

$X'=r\cos(\theta +\phi )=rcos\phi\ cos\theta\ -rsin\phi\ sin\theta$              ..........(3)

$Y'=r\sin(\theta +\phi )=rcos\phi\ sin\theta\ +rsin\phi\ cos\theta$             ............(4)

Substituting equation (1) & (2) in (3) & (4) respectively, we will get

$X'=Xcos\theta\ -Ysin\theta$

$Y'=Xsin\theta\ +Ycos\theta$

Representing the above equation in matrix form,

$\left [ X'Y' \right ]= \left [ XY \right ]\begin{bmatrix} cos\theta & sin\theta\\ -sin\theta& cos\theta \end{bmatrix}$

$P'= P . R$

Where R is the rotation matrix

$R=\begin{bmatrix} cos\theta &sin\theta \\ -sin\theta & cos\theta \end{bmatrix}$

The rotation angle can be positive and negative.

For positive rotation angle, we can use the above rotation matrix. However, for negative angle rotation, the matrix will change as shown below −

$R=\begin{bmatrix} cos(-\theta) &sin(-\theta) \\ -sin(-\theta) & cos(-\theta) \end{bmatrix}$

$R=\begin{bmatrix} cos\theta &-sin\theta \\ sin\theta& cos\theta \end{bmatrix}$

Comments

oh thankq..

one last question

positive rotation= anti clock wise...in the figure rt

negative rotation means ??

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off course clockwise rotation dear.

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@asu The prove given here is exactly what you want .I write it from a standard text book.So its correct.

Answer 2

It's a CG ques as far as I remember. I advice you to read Hearn and Baker for good reference on the above.  Yeah!!!. Hope it helps you. :)

Answer 3

just take a simple column vector x = [ 2  3]

if vector is rotated by zero degrees,then vector before transformation and after transformation is same.

   M.x=x

        where M is the transformation matrix.

 

put angle = zero degrees in options.

option-1 :

 M  =     1  0  

              0  1    

multiply matrix M with column vector x. (Mx )

 

after multiplication you are getting same vector =[2  3]    Mx=x

 

i.e, option 1 is satisfying