# Search results

1. ### Finding components of a vector

thanks ehild your post really helpful to me, now I got a more clear idea about mutually perpendicular unit vectors :), bye..
2. ### Finding components of a vector

Thanx a lot ehild :smile: bye..
3. ### Finding components of a vector

Thanx ehild, then what is my final answer would be?
4. ### Finding components of a vector

I found 3, 5 and 7 respectively for the components x1,x2 and x3. but I have a doubt about these values, I had to Find the components of d=(3,5,7) along the directions of u, v and w consider: u=1/3(2,2,-1) v=1/3(2,-1,2) w=1/3(-1,2,2), finally I got 3,5,7. this is really confusing me:confused:
5. ### Finding components of a vector

Oh...thank you very much ehild, I really appreciate your help:smile:, I have another problem, can you explain me little bit about mutually perpendicular unit vectors:smile:
6. ### Finding components of a vector

oops I've made a mistake when solving :smile:, is UU=1, then I got 3 for x1 :smile:
7. ### Finding components of a vector

I'm sorry, what did you mean by individual products, is it du,x1uu,x2uv and x3uw,if so x2uv and x3uw become zero,:redface:
8. ### Finding components of a vector

I used scalar product to solve it, du=x1uu+x2vu+x3wu (3,5,7)1/3(2,-2,-1)=x11/9(2,-2,-1)(2,-2,-1)+x21/3(2,-1,2)1/3(2,2,-1)+x31/3(-1,2,2)1/3(2,2,-1)
9. ### Finding components of a vector

after solving du=x1uu+x2vu+x3wu, I got 27 for x1 is this correct ehild?
10. ### Finding components of a vector

How do I suppose to calculate du? by using the scalar product??
11. ### Finding components of a vector

I don't understand, with which do I need to multiply??:confused: can you show me how to do that?
12. ### Finding components of a vector

please tell me how to apply scalar product, I know what is scalar product, but I don't know how to use it for find components
13. ### Finding components of a vector

hello! please someone help me,:smile: here is my question. Find the components of d=(3,5,7) along the directions of u, v and w consider: u=1/3(2,2,-1) v=1/3(2,-1,2) w=1/3(-1,2,2) I don't know where to start, I need some ideas to solve this thanx:smile:
14. ### Proving an equation using Properties of Determinants

So why don't u tell me the easyest way??:wink:
15. ### Proving an equation using Properties of Determinants

Hello guys I've asked to prove following equation on determinants, here it is; Using the properties of determinants & without expanding prove that, see attachment, I need to verify my answer can some one tell me whether is this correct or not?:smile:
16. ### Sumation of symmetric and skew symmetri metrices

Thank you I like Serena, I've found a formula to express a square matrix by using symmetric and skew symmetric matrices here it is; Let A be the given square matrix A can be uniquely expressed as sum of a symmetric matrix and a skew symmetric matrix, which is A =(A+A')/2 + (A-A')/2...
17. ### Sumation of symmetric and skew symmetri metrices

Thank you for fixing Latex vela :smile:, oh.. I think I've understood the question wrongly, so can you give me a hint, on how to do that in correct way :smile:
18. ### Sumation of symmetric and skew symmetri metrices

Express \left(\begin{array}{cccc} 6 & 1 & 5\\ -2 & -5 & 4\\ -3 & 3 & -1\ end{array} \right) as the sum of the symmetric and skew symmetric matrices. I did this following way Consider symmetric metric as "A" then; A = \left(\begin{array}{cccc} 6 & 1 & 5\\ 1 & -5 & 4\\ 5 & 4 & -1\ \end{array}...
19. ### Proveing Perpendicular of two vectors by applying Pythagoras rule

wow... thank you ehild:) your attachment really helpful for me, now I got a more clear idea about |c| = |b - a|.
20. ### Proveing Perpendicular of two vectors by applying Pythagoras rule

Thank you very much I like Serena. I got the idea, you are a really good helper:smile: hope to catch you again,:smile: bye...
21. ### Proveing Perpendicular of two vectors by applying Pythagoras rule

I know about addition and subtraction of vectors, what i wanted to understand is how did you get c=b-a instead of c=a+b, can you explain little bit more, what did you mean by, "c is the vector that starts from the end of vector a, and ends at the end of vector b. This means that c = b - a" ?
22. ### Proveing Perpendicular of two vectors by applying Pythagoras rule

Thank you so much I like Serena, finally I was able to proved that:smile:, can you explain how did you get the equation |a|2+|b|2=|b-a|2 more?:smile:
23. ### Proveing Perpendicular of two vectors by applying Pythagoras rule

vectors are realy hard lesson to me, so please show me how to prove that equation, I dont know where to start:confused: please help me!!
24. ### Proveing Perpendicular of two vectors by applying Pythagoras rule

according to the Pythagorean rule; for a Right triangle |a|2+|b|2=|c|2 can you give me a hint or something to solve my question properly, :smile:
25. ### Proveing Perpendicular of two vectors by applying Pythagoras rule

Given that a=(a1,a2,a3) and b=(b1,b2,b3) by applying the Pythagoras rule, Prove that a1b1+a2b2+a3b3=0 if a and b perpendicular The Scalar product a.b = |a||b|CosQ -------------(1) if two vectors are perpendicular; Q=90degrees then CosQ=0; from (1) a.b=0 (a1,a2,a3).(b1,b2,b3)=0...
26. ### Finding Potential Difference between two points homework

Oh.. thanks Femme_physics for your post:smile:, hope to get help again from both of you in future. bye..
27. ### Finding Potential Difference between two points homework

Thanks ehild for your advice :smile: I'll keep it on my mind, bye...
28. ### Finding Potential Difference between two points homework

Thank you so much ehild, I appreciate your help on this matter very much, once again thanks for helping me :smile: :smile:
29. ### Finding Potential Difference between two points homework

you mean to the final answer?? can you show me how to write the final answer :)
30. ### Finding Potential Difference between two points homework

Hi ehild, I solve the question, Total Resistance = R1+R3 = (120+2200)ohms consider current through Resister R1 and R3 is "I" then I=E1/(R1+R3) = 15/(120+2200) = 0.0065A At point "C" let's assume there to be zero potential 1) by going over R3 we lose -14.3V (0.0065*R3) 2) The E2 power supply...