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Magnitude of $$$\left\langle - \frac{\sin{\left(t \right)}}{3}, - \frac{\cos{\left(t \right)}}{3}, 0\right\rangle$$$
Your Input
Find the magnitude (length) of $$$\mathbf{\vec{u}} = \left\langle - \frac{\sin{\left(t \right)}}{3}, - \frac{\cos{\left(t \right)}}{3}, 0\right\rangle$$$.
Solution
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 $$$\left|{- \frac{\sin{\left(t \right)}}{3}}\right|^{2} + \left|{- \frac{\cos{\left(t \right)}}{3}}\right|^{2} + \left|{0}\right|^{2} = \frac{\sin^{2}{\left(t \right)}}{9} + \frac{\cos^{2}{\left(t \right)}}{9}$$$.
Therefore, the magnitude of the vector is $$$\mathbf{\left\lvert\vec{u}\right\rvert} = \sqrt{\frac{\sin^{2}{\left(t \right)}}{9} + \frac{\cos^{2}{\left(t \right)}}{9}} = \frac{1}{3}$$$.
Answer
The magnitude is $$$\frac{1}{3}\approx 0.333333333333333$$$A.