Problem 2 In this problem we will use the RIEMANN program to approximate the val
ID: 3167475 • Letter: P
Question
Explanation / Answer
Solutions to problem 13 (d)
First let us solve the ‘d’ part of the question.
All the steps are given for good understanding of the sequence of proof.
Given vectors a, b, c and d. Let us assume, for simplicity, that they are 2 dimensional.
ie. a = (p, q); b=(r, s); c=(u, v) and d = (x, y)
Then, by definition, a x b= (0, 0, ps-rq) , a vector normal to vectors a and b.
Similarly, c x d= (0, 0, uy-xv)
Hence, LHS = (a x b)-(c x d) = (ps-rq) (uy-xv) = psuy + rqxv – psxv - rquy
Now, simplify, the RHS of the ‘to be próved relation’
a. c = (p,q).(u,v) = pu + qv; b . d = rx + sy ; a .d = px + qy ; b .c = ru + sv
RHS = (a.c) (b.d)- (a.d)-(b.c) = [(pu + qv) (rx + sy)] – [(px + qy) ( (ru + sv)]
= purx + pusy + qvrx + qvsy – [ pxru + pxsv + qyru + qysv ]
= pusy + qvrx – pxsv – qyru
ie. LHS and RHS , on simplification, are reduced to the same quantity
viz. pusy + qvrx – pxsv – qyru . Hence LHS = RHS.
In other words, (a x b)-(c x d) = (a.c) (b.d)- (a.d)-(b.c).
This proves the required result.
We can take the vectors a,b,c and d to be 3 dimensional and the proof is exactly similar ,but with more computational effort.
Solution to (a), (b) and (c) is given separately.
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