What is the norm of #<1,-3,-2 >#?

1 Answer
Truong-Son N.
Apr 1, 2016

Let #vecv = << 1,-3,-2 >>#. The norm is written as

#\mathbf(|| vecv || = sqrt(vecvcdotvecv))#

#= sqrt(<< 1,-3,-2 >>cdot<< 1,-3,-2 >>)#

#= sqrt(1cdot1 + (-3)cdot(-3) + (-2)cdot(-2))#

#= sqrt(1^2 + (-3)^2 + (-2)^2)#

#= sqrt(1 + 9 + 4)#

#= color(blue)(sqrt(13))#

This tells us that the vector has a length of #sqrt(13) ~~ 3.61#. In fact, this is really a generalization of the Pythagorean Theorem in #RR^3#.

Given that information, can you find the norm of #vecw = << 1, 3, -2, 7 >>#?

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