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Question

# Let →p,→q,→r be three mutually perpendicular vectors of the same magnitude. If a vectors →x satisfies the equation →p×{(→x−→q)×→p}+→q×{(→x−→r)×→q}+→r×{(→x−→p)×→r}=→0 then →x is given by

A
12(p+q2r)
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B
12(p+q+r)
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C
13(p+q+r)
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D
13(2p+q+r)
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Solution

## The correct option is B 12(→p+→q+→r)→p×{(→x−→q)×→p}+→q×{(→x−→r)×→q}+→r×{(→x−→p)×→r}=→0⇒(→p.→p)(→x−→q)−[(→x−→q).→p]→p+(→q.→q)(→x−→r)−[(→x−→r).→q]→q+(→r.→r)(→x−→p)−[(→x−→p).→r]→r=→0⇒|→p|2(→x−→q)−(→p.→x)→p+|→q|2(→x−→r)−(→q.→x)→q+|→r|2(→x−→p)−(→r.→x)→r=→0Simplifying, we have |→p|2(3→x−→p−→q−→r)=(→p.→x)→p+(→q.→x)→q+(→r.→x)→rLets consider the expression written on RHS:(→p.→x)→p+(→q.→x)→q+(→r.→x)→rLet →x=a→p+b→q+→r→x.→p=a|→p|2→x.→q=b|→q|2=b|→p|2→x.→r=c|→r|2=c|→p|2So the expression RHS can be reduced toRHS =|→p|2(a→p+b→q+c→r)⇒ RHS =|→p|2→xHence the whole equation reduces to:|→p|2(3→x−→p−→q−→r)=|→p|2→xThus, we have →x as 12(→p+→q+→r)Hence, option B.

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