Magnitude of 5,2,3\left\langle 5, -2, 3\right\rangle

The calculator will find the magnitude (length, norm) of the vector 5,2,3\left\langle 5, -2, 3\right\rangle, with steps shown.
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Your Input

Find the magnitude (length) of u=5,2,3\mathbf{\vec{u}} = \left\langle 5, -2, 3\right\rangle.

Solution

The vector magnitude of a vector is given by the formula u=i=1nui2\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 52+22+32=38\left|{5}\right|^{2} + \left|{-2}\right|^{2} + \left|{3}\right|^{2} = 38.

Therefore, the magnitude of the vector is u=38\mathbf{\left\lvert\vec{u}\right\rvert} = \sqrt{38}.

Answer

The magnitude is 386.164414002968976\sqrt{38}\approx 6.164414002968976A.