Question

How do I find the magnitude and the direction angle for vector <12√2,-12√2>?

Answer 1

if i recall correctly maginitude is
\[\sqrt{a^2+b^2}\] just as in pythagoras

Answer 2

as for the angle, draw a picture and you will find instantly that it is \(-\frac{\pi}{4}\)

Answer 3

correct so I got 16 however that is not one of the answer choices so I guessed I was doing something wrong

Answer 4

\[\sqrt{(12\sqrt{2})^2+(12\sqrt{2})^2}\] i don't think it is 16

Answer 5

yeah i entered in my calc wrong

Answer 6

24 mag and -pi/4 is the solution then?

Answer 7

\[\sqrt{288+288}=\sqrt{576}=24\]

Answer 8

that is what i got, yes

Answer 9

ok great thank you so much

Answer 10

yw