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Let $\mathbf{u}=\left(u_{1}, u_{2}, u_{3}\right) .$ Explain why $\mathbf{u} \cdot \mathbf{u} \geq 0 .$ When is $\mathbf{u} \cdot \mathbf{u}=0 ?$

$(0,0,0)$

Calculus 3

Chapter 6

Orthogonality and Least Square

Section 1

Inner Product, Length, and Orthogonality

Vectors

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Campbell University

Baylor University

University of Michigan - Ann Arbor

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So we have a general vector. You want you to you three. And we want to see why the top product is always greater equal zero. So let's take you, don't you? So this is gonna be you want squared? Plus, you two squared closed, you three squared. But if we square number, this is greater or equal zero. This is greater equal zero, and this is greater equal zero. So if we add three things that are, um, at least zero, we're going to get something that is greater equals zero. So if everything, the only way to get zero is if each piece is zero So you want square, it has to be zero. Which means you want has to be zero, and so does you, too. And you three.

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