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 : $<1 , 4>$ , 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, $\begin{bmatrix} -3.96 \\ -1.13 \end{bmatrix}$

Answer 2

The matrix that represents the rotation of the vector 1,4 by 2pi/3 radians is : (B) $\left[\begin{array}{ccc}-3.96\\-1.13\\\end{array}\right]$

Meaning of Matrix

A matrix can be defined as a rectangular array of numbers table of numbers, symbols, or expressions that are arranged into column and rows.A matrix can take different forms which gave rise to the types of matrices.

Given that standard rotation matrix is expressed as :

$R = \left[\begin{array}{ccc}cos\beta &-sin\beta \\sin\beta &cos\beta \\\end{array}\right]$

therefore the matrix by rotation of $\frac{2\pi }{3}$

$\left[\begin{array}{ccc}x'\\y'\\\end{array}\right] = \left[\begin{array}{ccc}cos\beta &-sin\beta \\sin\beta &cos\beta \\\end{array}\right] \left[\begin{array}{ccc}x\\y\\\end{array}\right]$

substituting the value of $\beta = 120^o$

$\left[\begin{array}{ccc}x'\\y'\\\end{array}\right] = \left[\begin{array}{ccc}-0.5 &+4(-0.866) \\0.866 &+4(-0.5) \\\end{array}\right] \left[\begin{array}{ccc}\\\\\end{array}$

Therefore the rotation of the vector is

$\left[\begin{array}{ccc}-3.96\\-1.13\\\end{array}\right]$

In conclusion, The matrix that represents the rotation of the vector 1,4 by 2pi/3 radians is : (B).

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