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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[(xq)×p]+q×[(xr)×q]+r×[(xp)×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 andp.q=q.r=r.p=0} ...(i)
Given, p×{(xq)×p}+q×{(xr)×q}+r×{(xp)×r}=0
(p.p)(xq){p.(xq)}p+(q.q)(xr){q.(xr)}q+(r.r)(xp){r.(xp)}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
3x|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
3x|λ|2(p+q+r)|λ|2=|λ|22(p+q+r)
3x=12(p+q+r)+(p+q+r)
x=12(p+q+r)

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