Question

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)

Answer 1

do you know how to find the norm of a vector in \(\mathbb{R}^3\)?

Answer 2

\(||\mathbf{x}||:=\sqrt{x_1^2+x_2^2+x_3^2}\)

Answer 3

So norm of \(u\) is \(\sqrt{(-1)^2+2^2+2^2}=3\)

Answer 4

what is norm of \(w\) ?

Answer 5

sqrt(66) right?

Answer 6

correct

Answer 7

nope, your answer will be a real number.

Answer 8

by definition norms are real numbers

Answer 9

and notice it says "compute the following VALUES"

Answer 10

Do you know how to get \(-3u\) ?

Answer 11

Okay I understand now, thank you so much!

Answer 12

np

Answer 13

-3u is just -3(3) but the abs is 9.

Answer 14

\(-3u = 3i-6j-6k \)

Answer 15

\(||-3u||=\sqrt{3^2+6^2+6^2}\)

Answer 16

when we multiply a vector with a scalar, we multiply each component of the vector by that scalar.

Answer 17

so -3u is a vector not a scalar

Answer 18

so yes 9:)

Answer 19

because \(||-3u||=-3||u||\)

Answer 20

err

Answer 21

\(||-3u||=|-3|*||u||\)

Answer 22

you good?

Answer 23

okay thanks! what about the abs(4u-3v+w)

Answer 24

\(4u-3v+w = 4(-i+2j+2k)-3(-2i+j-3k)+(-5i+5j-4k)=\\-4i+8j+8k+6i-3j+9k-5i+5j-4k=-i+10j+13k\)
so we get \(\sqrt{(-1)^2+(10)^2+(13)^2}\)

Answer 25

So does the abs value have no effect?