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

The calculator will find the magnitude (length, norm) of the vector $$$\left\langle \cos{\left(t \right)}, - \sin{\left(t \right)}, 2 \sqrt{2}\right\rangle$$$, with steps shown.
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Your Input

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

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 $$$\left|{\cos{\left(t \right)}}\right|^{2} + \left|{- \sin{\left(t \right)}}\right|^{2} + \left|{2 \sqrt{2}}\right|^{2} = \sin^{2}{\left(t \right)} + \cos^{2}{\left(t \right)} + 8$$$.

Therefore, the magnitude of the vector is $$$\mathbf{\left\lvert\vec{u}\right\rvert} = \sqrt{\sin^{2}{\left(t \right)} + \cos^{2}{\left(t \right)} + 8} = 3$$$.

Answer

The magnitude is $$$3$$$A.