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Magnitude of $$$\left\langle 1, 3, 2 t\right\rangle$$$
Your Input
Find the magnitude (length) of $$$\mathbf{\vec{u}} = \left\langle 1, 3, 2 t\right\rangle$$$.
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 $$$\left|{1}\right|^{2} + \left|{3}\right|^{2} + \left|{2 t}\right|^{2} = 4 t^{2} + 10$$$.
Therefore, the magnitude of the vector is $$$\mathbf{\left\lvert\vec{u}\right\rvert} = \sqrt{4 t^{2} + 10}$$$.
Answer
The magnitude is $$$\sqrt{4 t^{2} + 10}\approx 3.162277660168379 \left(0.4 t^{2} + 1\right)^{0.5}$$$A.