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