Question

Which matrix represents the rotation of the vector 1,4 by 2pi/3 radians

Answer 1

Answer:

Option B is correct.

Rotation matrix = $\begin{bmatrix} -3.96 \\ -1.13 \end{bmatrix}$

Step-by-step explanation:

Given a vector : , rotation by $\frac{2\pi}{3}$ radian.

A rotation matrix is a matrix that is used to perform a rotation in Euclidean space.

The standard rotation matrix is given by;

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

Then, the matrix of rotation by $\frac{2\pi}{3}$ radian  is:

$\begin{bmatrix}x' \\ y'\end{bmatrix}$ = $\begin{bmatrix}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}$ $\begin{bmatrix}x \\ y\end{bmatrix}$

Then; substitute $\theta = 120^{\circ}$

$\begin{bmatrix}x' \\ y'\end{bmatrix}= \begin{bmatrix}\cos 120^{\circ} & -\sin 120^{\circ} \\ \sin 120^{\circ} & \cos 120^{\circ}\end{bmatrix}\begin{bmatrix}1 \\ 4 \end{bmatrix}$

or

$\begin{bmatrix}x' \\ y'\end{bmatrix}= \begin{bmatrix} -0.5 & -0.866 \\ 0.866 & -0.5 \end{bmatrix}\begin{bmatrix}1 \\ 4 \end{bmatrix}$

or

$\begin{bmatrix}x' \\ y'\end{bmatrix}= \begin{bmatrix} -0.5 +4(-0.866) \\ 0.866+4(-0.5)\end{bmatrix}$

Simplify:

$\begin{bmatrix}x' \\ y'\end{bmatrix} = \begin{bmatrix} -3.96 \\ -1.13 \end{bmatrix}$

Therefore, the rotation matrix of a given vector is, $\begin{bmatrix} -3.96 \\ -1.13 \end{bmatrix}$