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