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(D) Given equations are:$(i) x + y = a$$(ii) x 	imes y = b$$(iii) x \cdot a = 1$Taking the dot product of $(i)$ with $a$:$a \cdot (x + y) = a \cdot a

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None of these

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(D) Given equations are:$(i) x + y = a$$(ii) x 	imes y = b$$(iii) x \cdot a = 1$Taking the dot product of $(i)$ with $a$:$a \cdot (x + y) = a \cdot a$$a \cdot x + a \cdot y = |a|^2$$1 + a \cdot y = |a|^2 \implies a \cdot y = |a|^2 - 1$ $(iv)$Taking the cross product of $a$ with $(ii)$:$a 	imes (x 	imes y) = a 	imes b$$(a \cdot y)x - (a \cdot x)y = a 	imes b$Substituting $(iii)$ and $(iv)$:$(|a|^2 - 1)x - y = a 	imes b$ $(v)$From $(i)$,$y = a - x$. Substitute this into $(v)$:$(|a|^2 - 1)x - (a - x) = a 	imes b$$(|a|^2 - 1 + 1)x = a + (a 	imes b)$$|a|^2 x = a + (a 	imes b)$$x = \frac{a + (a 	imes b)}{|a|^2}$Then $y = a - x = a - \frac{a + (a 	imes b)}{|a|^2} = \frac{(|a|^2 - 1)a - (a 	imes b)}{|a|^2}$Since the calculated values do not match any of the given options,the correct choice is $(d)$.

Given the following simultaneous equations for vectors $x$ and $y$:
$(i) x + y = a$
$(ii) x \times y = b$
$(iii) x \cdot a = 1$
Then $x = ?, y = ?$

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Given the following simultaneous equations for vectors $x$ and $y$:$(i) x + y = a$$(ii) x \times y = b$$(iii) x \cdot a = 1$Then $x = ?, y = ?$