# Magnitude of $\left\langle - 6 t, 2, 6 t^{2}\right\rangle$

The calculator will find the magnitude (length, norm) of the vector $\left\langle - 6 t, 2, 6 t^{2}\right\rangle$, with steps shown.
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Find the magnitude (length) of $\mathbf{\vec{u}} = \left\langle - 6 t, 2, 6 t^{2}\right\rangle$.

### 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 $\left|{- 6 t}\right|^{2} + \left|{2}\right|^{2} + \left|{6 t^{2}}\right|^{2} = 36 t^{4} + 36 t^{2} + 4$.

Therefore, the magnitude of the vector is $\mathbf{\left\lvert\vec{u}\right\rvert} = \sqrt{36 t^{4} + 36 t^{2} + 4} = 2 \sqrt{9 t^{4} + 9 t^{2} + 1}$.

The magnitude is $2 \sqrt{9 t^{4} + 9 t^{2} + 1}\approx 6 \left(t^{4} + t^{2} + 0.111111111111111\right)^{0.5}$A.