Question
What transformation matrix would result in a 300 degrees counterclockwise rotation about the origin?
i cant type all of the options, but they all look like fractions inside brackets, some with square root signs.
i cant type all of the options, but they all look like fractions inside brackets, some with square root signs.
Asked by: USER3287
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225 Answers
Answer (225)
For rotation by an angle, θ, in the counter-clockwise direction about the origin, the transformation matrix is given by
[tex]\begin{pmatrix}cos\:\theta &-sin\:\theta \\ sin\:\theta &cos\:\theta \end{pmatrix}[/tex]
The given angle is θ=300°.
The values we need are
[tex]cos\left(300^{\circ} \right)=\frac{1}{2}[/tex]
[tex]sin\left(300^{\circ} \right)=-\frac{\sqrt{3}}{2}[/tex]
Substituting these into the given transformation matrix, we have
[tex]\begin{pmatrix}\frac{1}{2}&\frac{\sqrt{3}}{2}\\ -\frac{\sqrt{3}}{2}&\frac{1}{2}\end{pmatrix}[/tex]
[tex]\begin{pmatrix}cos\:\theta &-sin\:\theta \\ sin\:\theta &cos\:\theta \end{pmatrix}[/tex]
The given angle is θ=300°.
The values we need are
[tex]cos\left(300^{\circ} \right)=\frac{1}{2}[/tex]
[tex]sin\left(300^{\circ} \right)=-\frac{\sqrt{3}}{2}[/tex]
Substituting these into the given transformation matrix, we have
[tex]\begin{pmatrix}\frac{1}{2}&\frac{\sqrt{3}}{2}\\ -\frac{\sqrt{3}}{2}&\frac{1}{2}\end{pmatrix}[/tex]
the answer is B
hope this helps :)