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Find $$\mathbf{a}+\mathbf{b}, 2 \mathbf{a}+3 \mathbf{b},|\mathbf{a}|, and |\mathbf{a}-\mathbf{b}|$$$$\mathbf{a}=[2,-4,4], \quad \mathbf{b}=[0,2,-1]$$
$a+b=[2,-2,3]$$2 a+3 b=[4,-2,5]$$|a|=\sqrt{36}=6$$|a-b|=\sqrt{2^{2}+(-6)^{2}+(5)^{2}}=\sqrt{65}$
Calculus 3
Chapter 8
Vectors and Matrix Models
Section 2
Vectors
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Harvey Mudd College
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Okay, so for this question, we have a is equal to two negative for four. B is equal to zero to native 10 to negative one. So a plus B A plus B is equal to two plus zero. That's too negative. Four plus two. It's negative to four minus one that is three to a plus three feet to a plus. Three B is equal to, uh, well. It's four plus zero negative. Eight plus six eight minus three sore. Final answer is four negative to five magnitude of a is the square root off to swear plus four squared plus four square so that it's equal to the square root of 36 which is just equal to six. Then we have the magnitude of a minus bees. A minus B is equal to the magnitude of the square root of two square that's four plus six square. That's 36 plus five square that's 25. This is equal to the square root of 65
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