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)Since, →p,→q,→r are mutually perpendicular vectors of same magnitude, so let us consider |→p|=|→q|=|→r|=λ and→p.→q=→q.→r=→r.→p=0} ...(i) Given, →p×{(→x−→q)×→p}+→q×{(→x−→r)×→q}+→r×{(→x−→p)×→r}=→0 ⇒(→p.→p)(→x−→q)−{→p.(→x−→q)}→p+(→q.→q)(→x−→r)−{→q.(→x−→r)}→q+(→r.→r)(→x−→p)−{→r.(→x−→p)}→r=0 ⇒→x{(→p.→p)+(→q.→q)+(→r.→r)}−(→p.→p)→q−(→q.→q)r−(→r.→r)→p=(→x.→p)→p+(→x.→q)→q+(→x.→r)→r ⇒3→x|λ|2−(→p+→q+→r)|λ|2=(→x.→p)→p+(→x.→q)→q+(→x.→r)→r ...(ii) Taking dot of Eq. (ii) with →p we get 3(→x.→p)|λ|2−|λ|4=(→x.→p)|λ|2⇒=→x.→p=12|λ|2 Similarly, taking dot of Eq. (ii) with →q and →r respectively, we get →x.→q=|λ|22=→x.→r ∴ Eq. (ii) becomes 3→x|λ|2−(→p+→q+→r)|λ|2=|λ|22(→p+→q+→r) ⇒3→x=12(→p+→q+→r)+(→p+→q+→r) ⇒→x=12(→p+→q+→r)

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