Magnitude of $$$\left\langle 3 \sin^{2}{\left(t \right)} \cos{\left(t \right)}, - 3 \sin{\left(t \right)} \cos^{2}{\left(t \right)}, \sin{\left(2 t \right)}\right\rangle$$$

Find the magnitude (length) of $$$\mathbf{\vec{u}} = \left\langle 3 \sin^{2}{\left(t \right)} \cos{\left(t \right)}, - 3 \sin{\left(t \right)} \cos^{2}{\left(t \right)}, \sin{\left(2 t \right)}\right\rangle.$$$

The vector magnitude of a vector is given by the formula $$$\mathbf{\left\lvert\vec{u}\right\rvert} = \sqrt{\sum_{i=1}^{n} \left|{u_{i}}\right|^{2}}$$$.

The sum of squares of the absolute values of the coordinates is $$$\left|{3 \sin^{2}{\left(t \right)} \cos{\left(t \right)}}\right|^{2} + \left|{- 3 \sin{\left(t \right)} \cos^{2}{\left(t \right)}}\right|^{2} + \left|{\sin{\left(2 t \right)}}\right|^{2} = 9 \sin^{4}{\left(t \right)} \cos^{2}{\left(t \right)} + 9 \sin^{2}{\left(t \right)} \cos^{4}{\left(t \right)} + \sin^{2}{\left(2 t \right)}.$$$

Therefore, the magnitude of the vector is $$$\mathbf{\left\lvert\vec{u}\right\rvert} = \sqrt{9 \sin^{4}{\left(t \right)} \cos^{2}{\left(t \right)} + 9 \sin^{2}{\left(t \right)} \cos^{4}{\left(t \right)} + \sin^{2}{\left(2 t \right)}} = \frac{\sqrt{26 - 26 \cos{\left(4 t \right)}}}{4}.$$$

The magnitude is $$$\frac{\sqrt{26 - 26 \cos{\left(4 t \right)}}}{4}\approx 1.274754878398196 \left(1 - \cos{\left(4 t \right)}\right)^{0.5}$$$A.