# reciprocal vectors, Prob54

Discussion in 'Math' started by PG1995, Apr 15, 2012.

1. ### PG1995 Thread Starter Distinguished Member

Apr 15, 2011
813
6
Hi

Please have a look on the attachment and kindly help me with queries there. I'm sorry if answer to any of the queries is obvious. Right now, I'm very much exhausted and time-constrained. Thank you.

Regards
PG

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2. ### panic mode Senior Member

Oct 10, 2011
1,664
474
Q2: this is basic algebra:

start from
B(A*CxD) - A(B*CxD) = C(A*BxD) - D(A*BxC)

move the last term to the left (changes sign):
D(A*BxC) + B(A*CxD) - A(B*CxD) = C(A*BxD)

move second and third term to the right (they both change sign):
D(A*BxC) = -B(A*CxD) + A(B*CxD) + C(A*BxD)

reorder second and third term:
D(A*BxC) = A(B*CxD) - B(A*CxD) + C(A*BxD)

finally divide everything by bracket on the left in order to free D:
D(A*BxC)/(A*BxC) = [A(B*CxD) - B(A*CxD) + C(A*BxD)]/(A*BxC)

note (A*BxC)/(A*BxC) = 1

so we get

D*1 = [A(B*CxD) - B(A*CxD) + C(A*BxD)]/(A*BxC)
or
D= A(B*CxD)/(A*BxC) - B(A*CxD)/(A*BxC) + C(A*BxD)/(A*BxC)

3. ### PG1995 Thread Starter Distinguished Member

Apr 15, 2011
813
6
Thank you, panic mode, and sorry I have been little late with my thanks. Actually I was very much busy till yesterday.

Regards
PG