Question
Thu May 03, 2012 By: Student Learner

If p, q, r be 3 mutually perpendicular vectors...

Expert Reply
Wed May 09, 2012
Given that, p x [(x - q) x p] + q[(x - r) x q] + r[(x - p) x r] = 0
p x [(x x p - q x p] + q[x x q - r x q] + r[x x r - p x r] = 0
On simplifying, we get,
[(p.p) + (q.q) + (r.r)]x - [(p.p)q + (q.q)r + (r.r)p]  [(p.x)p + (q.x)q + (r.x)r] = 0  (1)
Since, |p| = |q| = |r| = p (say)
So, p.p + q.q + r.r = p2
p, q, r being mutually perpendicular, p.q + q.r + r.p = 0
Let x = x1p + x2q + x3r    (2)
Then, p.x = x1(p.p) + x2(p.q) + x3(p.r) = x1p2
Similarly, q.x = x2q2 = = x2p2 and r.x = x3p2
Since, x bar = 1/p2 [(p.x)p + (q.x)q + (r.x)r]
(p.x)p + (q.x)q + (r.x)r = p2 x bar
 
Putting the respective values in (1), we get,
3 p2 x bar - p2 [p + q + r] - p2 x bar = 0
This gives x bar = 1/2(p + q + r)
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