Question

what is the angle between -2.5+j4.33 and 4.33-j2.5??plz help..

Answer 1

use cross product or dot product

Answer 2

did you manage to work out something??

Answer 3

(-2.5+j4.33)(4.33-j2.5) like this??

Answer 4

take any dot product or cross ......do you know how to take these?

Answer 5

can you write how to do it?

Answer 6

ok
1. dot product is
\[a . b = \left| a \right| \left| b \right| \cos \theta \]
where theta is the angle b/w the two vectors
|a| = magnitude of the vector a

Answer 7

lets do it step by step
1. find the magnitude of both the vectors and tell me

Answer 8

24.9989 for the both vectors..

Answer 9

you also need to take squareroot of the expression for magnitude..isn`t it?

Answer 10

* sq. root 24.9989

Answer 11

cool
now plug in the values in the formula i gave you above and tell me

Answer 12

\[ \left|\sqrt{24.9989} \right| \times \left|\sqrt{24.9989} \right| \cos \theta\]

Answer 13

like that??

Answer 14

yes!! what about LHS

Answer 15

150?? that correct?

Answer 16

21.65 i suppose
it`s just simple multiplication of i terms and j terms and then their addition

Answer 17

how did you get 21.65?

Answer 18

-2.5 * 4.33 + 4.33 * -2.5

-21.65

Answer 19

150 is an angle i mean...sorry..

Answer 20

i a hurry !!
YES answer is correct
Cool

Answer 21

can you help me to express this sq. root of 15 + j 20??

Answer 22

yes just give me a moment

Answer 23

ok
consider 5(3 + j 4) = 5( 4 -1 + i 2 *2) = 5(2^2 + i^2 + 2*2*i) = 5 * (2 + i)^2

so squareroot is
\[\sqrt{5}(2 + i)\]

Answer 24

did you get it or any problem?

Answer 25

tnx..

Answer 26

:)