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Grootte van $$$\left\langle - \frac{\sqrt{6} \sin{\left(t + \frac{\pi}{4} \right)}}{3}, \frac{\sqrt{6} \cos{\left(t + \frac{\pi}{4} \right)}}{3}, 0\right\rangle$$$
Uw invoer
Bepaal de grootte (lengte) van $$$\mathbf{\vec{u}} = \left\langle - \frac{\sqrt{6} \sin{\left(t + \frac{\pi}{4} \right)}}{3}, \frac{\sqrt{6} \cos{\left(t + \frac{\pi}{4} \right)}}{3}, 0\right\rangle.$$$
Oplossing
De grootte van een vector wordt gegeven door de formule $$$\mathbf{\left\lvert\vec{u}\right\rvert} = \sqrt{\sum_{i=1}^{n} \left|{u_{i}}\right|^{2}}$$$.
De som van de kwadraten van de absolute waarden van de coördinaten is $$$\left|{- \frac{\sqrt{6} \sin{\left(t + \frac{\pi}{4} \right)}}{3}}\right|^{2} + \left|{\frac{\sqrt{6} \cos{\left(t + \frac{\pi}{4} \right)}}{3}}\right|^{2} + \left|{0}\right|^{2} = \frac{2 \sin^{2}{\left(t + \frac{\pi}{4} \right)}}{3} + \frac{2 \cos^{2}{\left(t + \frac{\pi}{4} \right)}}{3}.$$$
Daarom is de norm van de vector $$$\mathbf{\left\lvert\vec{u}\right\rvert} = \sqrt{\frac{2 \sin^{2}{\left(t + \frac{\pi}{4} \right)}}{3} + \frac{2 \cos^{2}{\left(t + \frac{\pi}{4} \right)}}{3}} = \frac{\sqrt{6}}{3}.$$$
Antwoord
De grootte is $$$\frac{\sqrt{6}}{3}\approx 0.816496580927726$$$A.