Magnitude of $\left\langle 6, -2, 0\right\rangle$

Find the magnitude (length) of $\mathbf{\vec{u}} = \left\langle 6, -2, 0\right\rangle$.

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 $\left|{6}\right|^{2} + \left|{-2}\right|^{2} + \left|{0}\right|^{2} = 40$.

Therefore, the magnitude of the vector is $\mathbf{\left\lvert\vec{u}\right\rvert} = \sqrt{40} = 2 \sqrt{10}$.

The magnitude is $2 \sqrt{10}\approx 6.324555320336759$A.