Question

can some body help me with this vector problem?
http://tinyurl.com/3wne3vx

Answer 1

ae is an addition of p+q; h is a fraction magnitude of AC

Answer 2

and its scalar so AE = hp + hq

AE = h(p+q)

Answer 3

is that other symbol a gradient symbol?

Answer 4

DE:EF is a ration of g:1...

Answer 5

DE = g(EF)

Answer 6

or just plain g = DE/EF

Answer 7

AF = k(p)

Answer 8

is g a magnitude?

Answer 9

that seems to be the best I can pull out if it....

Answer 10

i need help for part b

Answer 11

p*q = cos(t) |p||q|

Answer 12

(1/2)(2)(3) = 3

p*q = 3

Answer 13

how do you find k?

Answer 14

if you determined that g=1/k; then k = 1/g

Answer 15

if DF is perp to AC then:

DF * AF = 0

Answer 16

DF*AC = 0

Answer 17

given the magnitudes expressed as they are, that is helpful

Answer 18

so (-p+kq)*(p+q) = 0

Answer 19

|DF| = <kp>*<q>
---------
|kp||q|

and theres a place to put the cos(pi/3); just gotta recall it :)

Answer 20

its the law of cosines...

|DF|^2 = |k3|^2 + |2|^2 -2|k3||2| cos(pi/3)

Answer 21

|DF|^2 = 9k^2 + 4 -12k(1/2)

|DF|^2 = 9k^2 + 4 -6k

Answer 22

|DF|^2 = 9k^2 -6k + 4 dont know how much this is gonna be useful :)

Answer 23

angle b=120degrees or 2pi/3

Answer 24

<kp>-<q> = <DF>

Answer 25

k = p*q
-----
3

Answer 26

p*q = 3

|3k||2| (1/2) = kp*q

3k = k(p*q)

since DAB = pi/3, then q = <1,sqrt(3)>

Answer 27

p = <0,3>

Answer 28

right?

Answer 29

yeah

Answer 30

i get it now

Answer 31

im starting to get it :)

Answer 32

do you have any tips on how to solve vector because im going to have a very important test in 2 weeks?

Answer 33

nothing that you probably dont already know, ive been reading up on them for the past week to get a feel for them.