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Magnitude of $$$\left\langle - \frac{\cos{\left(t \right)}}{2}, 0, - \frac{\sin{\left(t \right)}}{2}\right\rangle$$$

The calculator will find the magnitude (length, norm) of the vector $$$\left\langle - \frac{\cos{\left(t \right)}}{2}, 0, - \frac{\sin{\left(t \right)}}{2}\right\rangle$$$, with steps shown.
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Your Input

Find the magnitude (length) of $$$\mathbf{\vec{u}} = \left\langle - \frac{\cos{\left(t \right)}}{2}, 0, - \frac{\sin{\left(t \right)}}{2}\right\rangle$$$.

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 $$$\left|{- \frac{\cos{\left(t \right)}}{2}}\right|^{2} + \left|{0}\right|^{2} + \left|{- \frac{\sin{\left(t \right)}}{2}}\right|^{2} = \frac{\sin^{2}{\left(t \right)}}{4} + \frac{\cos^{2}{\left(t \right)}}{4}$$$.

Therefore, the magnitude of the vector is $$$\mathbf{\left\lvert\vec{u}\right\rvert} = \sqrt{\frac{\sin^{2}{\left(t \right)}}{4} + \frac{\cos^{2}{\left(t \right)}}{4}} = \frac{1}{2}$$$.

Answer

The magnitude is $$$\frac{1}{2} = 0.5$$$A.