If x→−y→=a→, x→ ×y→=b→ and y→ . a→=−1, then

a→ (x→−y→)=a→ ⇒ a→. x→−a→ .y→=a→.a→⇒ a→. x→−a→ .y→= a→. x→+1=a2∴ a→. x→= a2−1∵ x→×y→=b→⇒ a→×(x→×y→)=a→×b→⇒ (a→.y→)x→−(a→. x→)y→=a→×b→⇒ −x→−(a2−1)y→=a→×b→⇒ x→=−(a2−1)y→−a→×b→∵ x→−y→=a→⇒ −(a2−1)y→−a→×b→−y→=a→⇒ y→=(b→×a→)−a→a2⇒x→=a→+y→=a→+(b→×a→−a→)a2

If x→−y→=a→, x→ ×y→=b→ and y→ . a→=−1, then

a→ (x→−y→)=a→ ⇒ a→. x→−a→ .y→=a→.a→⇒ a→. x→−a→ .y→= a→. x→+1=a2∴ a→. x→= a2−1∵ x→×y→=b→⇒ a→×(x→×y→)=a→×b→⇒ (a→.y→)x→−(a→. x→)y→=a→×b→⇒ −x→−(a2−1)y→=a→×b→⇒ x→=−(a2−1)y→−a→×b→∵ x→−y→=a→⇒ −(a2−1)y→−a→×b→−y→=a→⇒ y→=(b→×a→)−a→a2⇒x→=a→+y→=a→+(b→×a→−a→)a2

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If $\vec{x} - \vec{y} = \vec{a}, \vec{x} × \vec{y} = \vec{b} and \vec{y} . \vec{a} = - 1, then$

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$\vec{x} = \vec{a} + \frac{(\vec{b} × \vec{a} - \vec{a})}{a^{2}}$

$\vec{y} = \vec{a} + \frac{(\vec{b} × \vec{a}) - \vec{a}}{a^{2}}$

$x = a + \frac{(b × a) - a}{a^{2}}$

$\vec{y} = \vec{a} + \frac{(b × a - a)}{a^{2}}$

answer is A, B.

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Detailed Solution

$\begin{array}{l} \vec{a} (\vec{x} - \vec{y}) = \vec{a} \\ \Rightarrow \vec{a} . \vec{x} - \vec{a} . \vec{y} = \vec{a} . \vec{a} \\ \Rightarrow \vec{a} . \vec{x} - \vec{a} . \vec{y} = \vec{a} . \vec{x} + 1 = a^{2} \\ ∴ \vec{a} . \vec{x} = a^{2} - 1 \end{array}$

$\begin{array}{l} ∵ \vec{x} \times \vec{y} = \vec{b} \\ \Rightarrow \vec{a} \times (\vec{x} \times \vec{y}) = \vec{a} \times \vec{b} \\ \Rightarrow (\vec{a} . \vec{y}) \vec{x} - (\vec{a} . \vec{x}) \vec{y} = \vec{a} \times \vec{b} \\ \Rightarrow - \vec{x} - (a^{2} - 1) \vec{y} = \vec{a} \times \vec{b} \\ \Rightarrow \vec{x} = - (a^{2} - 1) \vec{y} - \vec{a} \times \vec{b} \end{array}$

$\begin{array}{l} ∵ \vec{x} - \vec{y} = \vec{a} \\ \Rightarrow - (a^{2} - 1) \vec{y} - \vec{a} \times \vec{b} - \vec{y} = \vec{a} \\ \Rightarrow \vec{y} = \frac{(\vec{b} \times \vec{a}) - \vec{a}}{a^{2}} \\ \Rightarrow \vec{x} = \vec{a} + \vec{y} = \vec{a} + \frac{(\vec{b} \times \vec{a} - \vec{a})}{a^{2}} \end{array}$