Question

Find unit vector

a-same direction as v
b-opposite direction of v

v=(2,2)

Answer 1

could you help me find the unit vector?

Answer 2

divide each component by the length of v to get the unit vector.

Answer 3

hmm...?

Answer 4

how would I do that?

Answer 5

what's the length or norm of v?

Answer 6

well, the coordinates are at (2,2)

Answer 7

I know that much

Answer 8

\[||v|| =\sqrt{2^2+2^2}=2\sqrt{2}\]
unit vector has length of 1 =>
\[\frac{v}{||v||}\]

Answer 9

wait

Answer 10

how did you find out that the unit vector has a length of 1 based on just achieving the result 2 and the square root of 2?

Answer 11

a unit vector is defined to have a lenght of 1... similar to the unit circle: defined as having a radius of 1.

Answer 12

ahh

Answer 13

gotta know your defs for math! it's a language

Answer 14

what's the next step?

Answer 15

I got llvll=2 and the square root of 2 and the unit vector has a length of 1

Answer 16

what next?

Answer 17

sorry if I am clueless, I just have a different precalculus book than my professor

Answer 18

once I grasp the logic of the problem, it is like cracking a lock open

Answer 19

to find the unit vector which points in the direction of v, simply divide v by its norm. this involves dividing each compnent of v by its norm.
\[\vec{u}=\frac{\vec{v}}{||\vec{v}||}=\left(\frac{2}{2\sqrt{2}},\,\frac{2}{2\sqrt{2}}\right) \]

Answer 20

ahh

Answer 21

to find the one pointing in the opposite direction, simply multiply the unit vector by -1.

Also, you should simplify the previous.

Answer 22

so the opposite direction would be \[(-2/2\sqrt{2}, -2/2\sqrt{2}\]

Answer 23

yes, but simplify. no radicals on the bottom.

Answer 24

(-2/2 square root of 2, -2/2 square root of 2

Answer 25

-2 square root of 2/4?

Answer 26

learn to use the equation editor... it makes things much clearer!

\[\vec{u}=\left(\frac{\sqrt{2}}{2},\,\frac{\sqrt{2}}{2}\right)\]
\[-\vec{u}=\left(-\frac{\sqrt{2}}{2},\,-\frac{\sqrt{2}}{2}\right)\]

Answer 27

ty

Answer 28

you're welcome