If x→+c→×y→=a→ and y→+c→×x→=b→, where C→ is a non–zero vector, then which of the following is not correct?

x→+c→×y→=a→…….(i)⇒y→+c→×x→=b→……(ii) Taking cross with c→, we have c→×y→+c→×(c→×x→)=c→×b→⇒(a→−x→)+(c→⋅x→)c→−(c→⋅c→)x→=c→×b→ Also x→+c→×y→=a→⇒c→⋅x→+c→⋅(c→×y→)=c→⋅a→ or c→⋅x→+0=c→⋅a→ or c→⋅x→=c→⋅a→⇒a→−x→+(c→⋅a→)c→−(c→.c→)x→=c→×b→ or x→(1+(c→⋅c→))=b→×c→+a→+(c→⋅a→)⋅c→ or x→=b→×c→+a→+(c→⋅a→)⋅c→1+c→⋅c→ Similarly on taking cross product of Eq. (i), we find y→=a→×c→+b→+(c→⋅b→)⋅c→1+c→⋅c→

If x→+c→×y→=a→ and y→+c→×x→=b→, where C→ is a non–zero vector, then which of the following is not correct?

x→+c→×y→=a→…….(i)⇒y→+c→×x→=b→……(ii) Taking cross with c→, we have c→×y→+c→×(c→×x→)=c→×b→⇒(a→−x→)+(c→⋅x→)c→−(c→⋅c→)x→=c→×b→ Also x→+c→×y→=a→⇒c→⋅x→+c→⋅(c→×y→)=c→⋅a→ or c→⋅x→+0=c→⋅a→ or c→⋅x→=c→⋅a→⇒a→−x→+(c→⋅a→)c→−(c→.c→)x→=c→×b→ or x→(1+(c→⋅c→))=b→×c→+a→+(c→⋅a→)⋅c→ or x→=b→×c→+a→+(c→⋅a→)⋅c→1+c→⋅c→ Similarly on taking cross product of Eq. (i), we find y→=a→×c→+b→+(c→⋅b→)⋅c→1+c→⋅c→

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If $\vec{x} + \vec{c} \times \vec{y} = \vec{a}$ and $\vec{y} + \vec{c} \times \vec{x} = \vec{b},$ where $\vec{C}$ is a non–zero vector, then which of the following is not correct?

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$\vec{x} = \frac{\vec{c} \times \vec{b} + \vec{b} + (\vec{c} \cdot \vec{a}) \vec{c}}{1 + \vec{c} \cdot \vec{c}}$

$\vec{y} = \frac{\vec{a} \times \vec{c} + \vec{b} + (\vec{c} \cdot \vec{b}) \vec{c}}{1 + \vec{c} \cdot \vec{c}}$

$\vec{x} = \frac{\vec{b} \times \vec{c} + \vec{a} + (\vec{c} \cdot \vec{a}) \vec{c}}{1 + \vec{c} \cdot \vec{c}}$

$\bar{x} = \frac{\bar{c} \times \bar{b} - \bar{a} (\bar{a} \cdot \bar{c}) \bar{c}}{- (1 + \bar{c} \cdot \bar{c})}$

answer is B.

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$\vec{x} + \vec{c} \times \vec{y} = \vec{a} \dots \dots . (i) \Rightarrow \vec{y} + \vec{c} \times \vec{x} = \vec{b} \dots \dots (i i) Taking$

$cross with \vec{c}, we have \vec{c} \times \vec{y} + \vec{c} \times (\vec{c} \times \vec{x}) = \vec{c} \times \vec{b}$

$\Rightarrow (\vec{a} - \vec{x}) + (\vec{c} \cdot \vec{x}) \vec{c} - (\vec{c} \cdot \vec{c}) \vec{x} = \vec{c} \times \vec{b}$

$Also \vec{x} + \vec{c} \times \vec{y} = \vec{a} \Rightarrow \vec{c} \cdot \vec{x} + \vec{c} \cdot (\vec{c} \times \vec{y}) = \vec{c} \cdot \vec{a}$

$or \vec{c} \cdot \vec{x} + 0 = \vec{c} \cdot \vec{a} or \vec{c} \cdot \vec{x} = \vec{c} \cdot \vec{a}$

$\Rightarrow \vec{a} - \vec{x} + (\vec{c} \cdot \vec{a}) \vec{c} - (\vec{c} . \vec{c}) \vec{x} = \vec{c} \times \vec{b}$

$or \vec{x} (1 + (\vec{c} \cdot \vec{c})) = \vec{b} \times \vec{c} + \vec{a} + (\vec{c} \cdot \vec{a}) \cdot \vec{c}$

$or \vec{x} = \frac{\vec{b} \times \vec{c} + \vec{a} + (\vec{c} \cdot \vec{a}) \cdot \vec{c}}{1 + \vec{c} \cdot \vec{c}} Similarly on taking cross$

product of Eq. (i), we find $\vec{y} = \frac{\vec{a} \times \vec{c} + \vec{b} + (\vec{c} \cdot \vec{b}) \cdot \vec{c}}{1 + \vec{c} \cdot \vec{c}}$

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If x→+c→×y→=a→ and y→+c→×x→=b→, where C→ is a non–zero vector, then which of the following is not correct?

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If $\vec{x} + \vec{c} \times \vec{y} = \vec{a}$ and $\vec{y} + \vec{c} \times \vec{x} = \vec{b},$ where $\vec{C}$ is a non–zero vector, then which of the following is not correct?