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Find the distance between $\mathbf{u}=\left[\begin{array}{r}{0} \\ {-5} \\ {2}\end{array}\right]$ and $\mathbf{z}=\left[\begin{array}{r}{-4} \\ {-1} \\ {8}\end{array}\right]$

2$\sqrt{17}$

Calculus 3

Chapter 6

Orthogonality and Least Square

Section 1

Inner Product, Length, and Orthogonality

Vectors

Johns Hopkins University

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Harvey Mudd College

University of Michigan - Ann Arbor

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So we want to find the distance between these two vectors. So in order to do that, we take the norm of their difference. So if we want to subtract thes two vectors, we subtract on them component wise. So we have zero minus a negative, for which is for negative five minus negative. One is negative for on and two minus eight is negative. Six. So the normal this vector is equal to the square root of four squared plus negative four squared plus negative. Six squared. So we get 16 full 16 close 36 which gives us 32 plus 36 which gives us 68. And we can simplify this a little bit. Ah, 68 breaks down into four times 17 so have yet to square it's of 70.

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