Recent content by harshakantha

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

Thanx a lot ehild :smile: bye..

Thanx ehild, then what is my final answer would be?

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:

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:

oops I've made a mistake when solving :smile:, is UU=1, then I got 3 for x1 :smile:

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:

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)

after solving du=x1uu+x2vu+x3wu, I got 27 for x1 is this correct ehild?

How do I suppose to calculate du? by using the scalar product??

I don't understand, with which do I need to multiply??:confused: can you show me how to do that?

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

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:

So why don't u tell me the easyest way??:wink:

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: