a and b are two vectors such that |a|=sqrt{3} | vedclass

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(C) Given $x 	imes a = b$. Taking the cross product with $a$ on both sides:$a 	imes (x 	imes a) = a 	imes b$Using the vector triple product identi

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$\frac{1}{2}[(x \cdot a) a + b 	imes a]$

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(C) Given $x 	imes a = b$. Taking the cross product with $a$ on both sides:$a 	imes (x 	imes a) = a 	imes b$Using the vector triple product identity $a 	imes (b 	imes c) = (a \cdot c)b - (a \cdot b)c$:$(a \cdot a)x - (a \cdot x)a = a 	imes b$Since $|a| = \sqrt{3}$,we have $a \cdot a = |a|^2 = 3$:$3x - (a \cdot x)a = a 	imes b$$3x = (a \cdot x)a + a 	imes b$$x = \frac{1}{3}[(a \cdot x)a + a 	imes b]$Since $a 	imes b = -(b 	imes a)$,we can write:$x = \frac{1}{3}[(a \cdot x)a - b 	imes a]$Given the options provided,the structure matches the form involving $(x \cdot a)a$ and $b 	imes a$.

$a$ and $b$ are two vectors such that $|a|=\sqrt{3}$ and $|b|=\sqrt{2}$. If $x$ is a unit vector satisfying $x \times a = b$,then $x$ is equal to:

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