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Question

Let →p,→q,→r be three mutually perpendicular vectors of the same magnitude. If a vector →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+qr)
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Solution

The correct option is B 12(→p+→q+→r)We have →p×{→x−→q×→p}+→q×{→x−→r×→q}+→r×{→x−→p×→r}=→0⇒(→p.→p)(→x−→p)−{→p(→x−→q)}→p+(→q.→q)(→x−→r)−{→q.(→x−→r)}→q+(→r.→r)(→x−→p)−{→r.(→x−→p)}→r=0⇒(∣∣→p∣∣2=∣∣→q∣∣2+∣∣→r∣∣2)→x−(∣∣→p∣∣2→q+∣∣→q∣∣2→r+∣∣→r∣∣2→p)−[(→p.→x)→p+(→q.→x)→q+(→r.→x)→r]=0⇒3λ2→x−λ2(→p+→q+→r)−{(→p.→x)→p+(→r.→x)→q+(→r.→x)→r}=0 ...(1)where λ=∣∣→p∣∣=∣∣→q∣∣=∣∣→r∣∣Taking dot product with →p, we get 3λ2(→x.→p)−λ2∣∣→p∣∣2=0[∵→p.→q=→p.→r=0]⇒3λ2(→x.→p)−λ4−λ2(→p.→x)=0⇒→p.→x=λ22Similarly , →q.→x=λ22 and →r.→x=λ22.Substituting these vales in (1), we get3λ2→x.λ2(→p+→q+→r)−λ22(→p+→q+→r)=→0⇒→x=12(→p+→q+→r)

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