Question

Does anyone know this? It says my calculations are off by 10%

Vector B (with arrow above) has x, y, and z components of 6.00, 8.00, and 5.00 units, respectively. Calculate the magnitude of B (with arrow above).

_11.18_

Calulcate the angle that makes with the x axis.
____°
Calculate the angle that makes with the y axis.
____°
Calculate the angle that makes with the z axis.
____°

Answer 1

how much hw you got?

Answer 2

last one lol

Answer 3

but she gives us a lot with stuff we havent learned yet lol its why im on here

Answer 4

11.1803

Answer 5

i got that lol

Answer 6

i need the rest

Answer 7

or 5 sqrt(5)

Answer 8

have you learned dot products yet?

Answer 9

for angles i did arc tan of each of these
6/8
8/5
6/5

it told me i was wrong

Answer 10

no

Answer 11

acos

Answer 12

e.g acos(6/11,18)

Answer 13

{6,8,5} is your vector

{1,0,0} is x axis
\[\cos ^{-1}\left(\frac{A.B}{|A| |B|}\right)=\theta\]

\[\text{ArcCos}\left[\frac{6}{5 \sqrt{5}}\right]\]=58 degree

Answer 14

you can use the \vec command to make the arrow

\[\vec{B}\]

Answer 15

57.5437 to be more accurate

Answer 16

ok so thats for x

Answer 17

so for y is it arc sin?

Answer 18

no,
\[\left[\text{ArcCos}\left[\frac{8}{5 \sqrt{5}}\right]\right.\]
=44.31

Answer 19

so its a triangle right?
so its 180-44.31-57.54?

Answer 20

no,
\[\text{ArcCos}\left[\frac{5}{5 \sqrt{5}}\right]\]
for z

which is
63.43

Answer 21

i feel so dumb..i wish my teacher was better =(
thanks for all your help today

Answer 22

it is tough to do 3-dim. But if you focus on just one axis, say z.
you have a line going straight up (z) and another slanting off (B)
B is the hypotenuse

Answer 23

did that check out?

Answer 24

the cosine of the angle is z component/hypotenuse

Answer 25

yea it all was correct! thanks!!! =)

Answer 26

nice, so basically
\[\text{ArcCos}\left[\frac{\text{Component}}{\text{Magnitude}}\right]\]