Need help with Vector Calculus problems due tomorrow!

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Help is requested for solving two vector calculus problems due tomorrow, focusing on determinants and vector operations. The discussion emphasizes the need to expand determinants for cross products and perform dot products to prove equalities. Key points include the properties of the cross product, such as its non-associativity and the scalar triple product's relation to the volume of a parallelepiped. Participants clarify the meanings of LHS (Left Hand Side) and RHS (Right Hand Side) in the context of the problems. Overall, the thread provides guidance on how to approach and solve the vector calculus problems effectively.
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Vectorcalculus! Help Needed B4 Tomorrow!

NOTE: bold characters are vectors

Could somebody please help me to show that:

http://img259.imageshack.us/img259/8651/problem6dx.gif

and to proof:

http://img403.imageshack.us/img403/6641/problem26dw.gif

I haven't had any explanations on vectors & determinants and my teacher asks me to solve this problem... I know how determinants work, but I don't know how to translate this to the vector problem above. Please help me! I need to finish this by tomorrow (just heard it today)!
 
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Find LHS and RHS indepedently and then prove that you are reaching the same result. If you have a bit more problems please show how you argued so that I can tell what is wrong in your arguement.
 
For the first problem, expand the determinant of the matrix for the cross product, for example, AxB is the determinant of:
\left(\begin{array}{ccc}i&b_x&c_x\\j&b_y&c_y\\k&b_z&c_z\end{array}\right)
expand that determinant, and then do the dot product of it, and it is quite obviously the determinant of the matrix shown, and obviously C.(AxB) is true, since they relate to the properties of vectors.
* The i,j and k in the matrix represent the components of the vectors.

Did you know that Ax(BxC) = (AxB)xC, etc. Do you thing you can do that one now.

Hope that helps you with most of it.:biggrin:
 
firs tdo the determinant.
change all the vectors to their components. then do each operation.

cross AxB=(aybz-byaz,azbx-bzax,axby-bxay)
Dot A.B=(axbx+ayby+azbz);
 
finchie_88 said:
Did you know that Ax(BxC) = (AxB)xC, etc. Do you thing you can do that one now.

The cross-product is not associative!
Look at the OP's identity (1)... the Jacobi Identity. Bringing the last term on the LHS of (1) onto the other side is your RHS.
 
I managed to figure out why A . (B x C) has the give determinant, but why do C . (A x B) and B . (C x A) have the same deteriminant? The drawn determinant is a det(ABC) and the other two would be det(CAB) and det(BCA) ...
 
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What is LHS and RHS actually?!
 
As to why the three are equal, this is because the parallelepiped created by A, B and C is the absolute value of this (it is called the scalar triple product).

For the second set, first show #2 by xpanding it out. #1 follows from that.
 
Could you explain a bit more ... I don't get what you're saying...
 
  • #10
Lisa... said:
What is LHS and RHS actually?!

I may be missinterpreting you, but LHS means Left Hand Side, and RHS is the Right Hand Side.

Hope that helps,

Gareth
 
  • #11
look on line for the image of parallelpipe...best website choice is probably mathworld.com
 
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