Magnitude of $\left\langle \sqrt{2}, -1, 1\right\rangle$

Find the magnitude (length) of $\mathbf{\vec{u}} = \left\langle \sqrt{2}, -1, 1\right\rangle$.

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 $\left|{\sqrt{2}}\right|^{2} + \left|{-1}\right|^{2} + \left|{1}\right|^{2} = 4$.

Therefore, the magnitude of the vector is $\mathbf{\left\lvert\vec{u}\right\rvert} = \sqrt{4} = 2$.

The magnitude is $2$A.