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

Let |p|=|q|=|r|=K. Let a,b,c be unit vectors along p, q, r respectively. Clearly p,q,r are mutually perpendicular vectors, so any vector x can be written as a1a+a2b+a3cp×((x−q)×p)=(p⋅p)(x−q)−(p⋅(x−q))p=K2(x−q)−(p⋅x)p(p⋅q=0)=K2(x−q)−Ka⋅a1a+a2b+a3cKa=K2x−q−a1aSimilarly q×((x−r)×q)=K2x−r−a2band r×((x−p)×r)=K2x−p−a3cAccording to the given condition K2x−q−a1a+x−r−a2b+x−p−a3c=0⇒ 3x−(p+q+r)−a1a+a2b+a3c=0⇒ [2x−(p+q+r)]=0⇒x=(1/2)(p+q+r)

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

Let |p|=|q|=|r|=K. Let a,b,c be unit vectors along p, q, r respectively. Clearly p,q,r are mutually perpendicular vectors, so any vector x can be written as a1a+a2b+a3cp×((x−q)×p)=(p⋅p)(x−q)−(p⋅(x−q))p=K2(x−q)−(p⋅x)p(p⋅q=0)=K2(x−q)−Ka⋅a1a+a2b+a3cKa=K2x−q−a1aSimilarly q×((x−r)×q)=K2x−r−a2band r×((x−p)×r)=K2x−p−a3cAccording to the given condition K2x−q−a1a+x−r−a2b+x−p−a3c=0⇒ 3x−(p+q+r)−a1a+a2b+a3c=0⇒ [2x−(p+q+r)]=0⇒x=(1/2)(p+q+r)

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Let $p, q, r$ be three mutually perpendicular vectors of the same magnitude. If a vector $\vec{x}$ satisfies the equation
$\vec{p} \times ((\vec{x} - \vec{q}) \times \vec{p}) + \vec{q} \times ((\vec{x} - \vec{r}) \times \vec{q}) + \vec{r} \times ((\vec{x} - \vec{p}) \times \vec{r}) = \vec{0}$ then $\vec{x}$ is given by

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$(1 / 2) (\vec{p} + \vec{q} - 2 \vec{r})$

$(1 / 2) (\vec{p} + \vec{q} + \vec{r})$

$(1 / 3) (2 \vec{p} + \vec{q} - \vec{r})$

$(1 / 3) \vec{(p} + \vec{q} + \vec{r})$

answer is B.

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Detailed Solution

$Let | p | = | q | = | r | = K . Let a, b, c$ be unit vectors along p, q, r respectively. Clearly $p, q, r$ are mutually perpendicular vectors, so any vector $x$ can be written as $a_{1} a + a_{2} b + a_{3} c$

$p \times ((x - q) \times p) = (p \cdot p) (x - q) - (p \cdot (x - q)) p$

$\begin{aligned} = K^{2} (x - q) - (p \cdot x) p (p \cdot q = 0) \\ = K^{2} (x - q) - K a \cdot (a_{1} a + a_{2} b + a_{3} c) K a \\ = K^{2} (x - q - a_{1} a) \end{aligned}$

Similarly $q \times ((x - r) \times q) = K^{2} (x - r - a_{2} b)$

and $r \times ((x - p) \times r) = K^{2} (x - p - a_{3} c)$

According to the given condition

$K^{2} (x - q - a_{1} a + x - r - a_{2} b + x - p - a_{3} c) = 0$

$\begin{array}{l} \Rightarrow \{3 x - (p + q + r) - (a_{1} a + a_{2} b + a_{3} c)\} = 0 \\ \Rightarrow [2 x - (p + q + r)] = 0 \\ \Rightarrow x = (1 / 2) (p + q + r) \end{array}$