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Magnitude of $$$\left\langle 1, - \frac{12}{25}, \frac{9}{25}\right\rangle$$$
Your Input
Find the magnitude (length) of $$$\mathbf{\vec{u}} = \left\langle 1, - \frac{12}{25}, \frac{9}{25}\right\rangle$$$.
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 $$$\left|{1}\right|^{2} + \left|{- \frac{12}{25}}\right|^{2} + \left|{\frac{9}{25}}\right|^{2} = \frac{34}{25}$$$.
Therefore, the magnitude of the vector is $$$\mathbf{\left\lvert\vec{u}\right\rvert} = \sqrt{\frac{34}{25}} = \frac{\sqrt{34}}{5}$$$.
Answer
The magnitude is $$$\frac{\sqrt{34}}{5}\approx 1.16619037896906$$$A.