Question

Rotate 30 degree around vector 2i+1j+3k

Answer 1

Given

Angle $\[ = 30^\circ \]$

Vector $\[ = 2\hat i + 1\hat j + 3\hat k\]$

Solution,

Consider the vector is rotated in anticlockwise direction,

$\[ = \left( {\begin{array}{*{20}{c}}{\cos 30^\circ }&{ - \sin 30^\circ }&0\\{\sin 30^\circ }&{\cos 30^\circ }&0\\0&0&1\end{array}} \right)\left( {\begin{array}{*{20}{c}}2\\1\\3\end{array}} \right)\]$

Put the value of trigonometric terms,

$\[ = \left( {\begin{array}{*{20}{c}}{\frac{{\sqrt 3 }}{2}}&{ - \frac{1}{2}}&0\\{\frac{1}{2}}&{\frac{{\sqrt 3 }}{2}}&0\\0&0&1\end{array}} \right)\left( {\begin{array}{*{20}{c}}2\\1\\3\end{array}} \right)\]$

Multiply the matrix,

$\[ = \left( {\begin{array}{*{20}{c}}{\frac{{\sqrt 3 }}{2} \times 2 - \frac{1}{2} \times 1 + 0 \times 3}\\{\frac{1}{2} \times 2 + \frac{{\sqrt 3 }}{2} \times 1 + 0 \times 3}\\{0 \times 2 + 0 \times 1 + 1 \times 3}\end{array}} \right)\]$

Simplify it

$\[ = \left( {\begin{array}{*{20}{c}}{\frac{{2\sqrt 3 - 1}}{2}}\\{\frac{{2 + \sqrt 3 }}{2}}\\3\end{array}} \right)\]$

New vector,

$\[ = \left( {\frac{{2\sqrt 3 - 1}}{2}} \right)\hat i + \left( {\frac{{2 + \sqrt 3 }}{2}} \right)\hat j + 3\hat k\]$

Hence, the new vector after rotating by $\[30^\circ \]$ is$\[\left( {\frac{{2\sqrt 3 - 1}}{2}} \right)\hat i + \left( {\frac{{2 + \sqrt 3 }}{2}} \right)\hat j + 3\hat k\]$.