Magnitude of $\left\langle 1, - \frac{12}{25}, \frac{9}{25}\right\rangle$

Find the magnitude (length) of $\mathbf{\vec{u}} = \left\langle 1, - \frac{12}{25}, \frac{9}{25}\right\rangle$.

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 $\left|{1}\right|^{2} + \left|{- \frac{12}{25}}\right|^{2} + \left|{\frac{9}{25}}\right|^{2} = \frac{34}{25}$.

Therefore, the magnitude of the vector is $\mathbf{\left\lvert\vec{u}\right\rvert} = \sqrt{\frac{34}{25}} = \frac{\sqrt{34}}{5}$.

The magnitude is $\frac{\sqrt{34}}{5}\approx 1.16619037896906$A.