Suppose vector u=-i+2j+2k, vector v=-2i+j-3k and vector w=-5i+5j-4k. Compute the following values: Abs(vector u)+ abs(vector v) Abs(vector -3u)+ 3abs(vector v)
do you know how to find the norm of a vector in \(\mathbb{R}^3\)?
\(||\mathbf{x}||:=\sqrt{x_1^2+x_2^2+x_3^2}\)
So norm of \(u\) is \(\sqrt{(-1)^2+2^2+2^2}=3\)
what is norm of \(w\) ?
sqrt(66) right?
correct
nope, your answer will be a real number.
by definition norms are real numbers
and notice it says "compute the following VALUES"
Do you know how to get \(-3u\) ?
Okay I understand now, thank you so much!
np
-3u is just -3(3) but the abs is 9.
\(-3u = 3i-6j-6k \)
\(||-3u||=\sqrt{3^2+6^2+6^2}\)
when we multiply a vector with a scalar, we multiply each component of the vector by that scalar.
so -3u is a vector not a scalar
so yes 9:)
because \(||-3u||=-3||u||\)
err
\(||-3u||=|-3|*||u||\)
you good?
okay thanks! what about the abs(4u-3v+w)
\(4u-3v+w = 4(-i+2j+2k)-3(-2i+j-3k)+(-5i+5j-4k)=\\-4i+8j+8k+6i-3j+9k-5i+5j-4k=-i+10j+13k\) so we get \(\sqrt{(-1)^2+(10)^2+(13)^2}\)
So does the abs value have no effect?
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