Magnitude of $$$\left\langle \sqrt{2} e^{t} \cos{\left(t + \frac{\pi}{4} \right)}, \sqrt{2} e^{t} \sin{\left(t + \frac{\pi}{4} \right)}, e^{t}\right\rangle$$$

Find the magnitude (length) of $$$\mathbf{\vec{u}} = \left\langle \sqrt{2} e^{t} \cos{\left(t + \frac{\pi}{4} \right)}, \sqrt{2} e^{t} \sin{\left(t + \frac{\pi}{4} \right)}, e^{t}\right\rangle.$$$

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 $$$\left|{\sqrt{2} e^{t} \cos{\left(t + \frac{\pi}{4} \right)}}\right|^{2} + \left|{\sqrt{2} e^{t} \sin{\left(t + \frac{\pi}{4} \right)}}\right|^{2} + \left|{e^{t}}\right|^{2} = 2 e^{2 t} \sin^{2}{\left(t + \frac{\pi}{4} \right)} + 2 e^{2 t} \cos^{2}{\left(t + \frac{\pi}{4} \right)} + e^{2 t}.$$$

Therefore, the magnitude of the vector is $$$\mathbf{\left\lvert\vec{u}\right\rvert} = \sqrt{2 e^{2 t} \sin^{2}{\left(t + \frac{\pi}{4} \right)} + 2 e^{2 t} \cos^{2}{\left(t + \frac{\pi}{4} \right)} + e^{2 t}} = \sqrt{3} e^{t}.$$$

The magnitude is $$$\sqrt{3} e^{t}\approx 1.732050807568877 e^{t}$$$A.