If a and b are unit vectors,then the vector ( | vedclass

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(A) Given that $a$ and $b$ are unit vectors,so $|a| = 1$ and $|b| = 1$,which implies $a \cdot a = 1$ and $b \cdot b = 1$. Expanding the expression $(a

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$a-b$

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(A) Given that $a$ and $b$ are unit vectors,so $|a| = 1$ and $|b| = 1$,which implies $a \cdot a = 1$ and $b \cdot b = 1$. Expanding the expression $(a+b) 	imes (a 	imes b)$ using the distributive property of the cross product: $(a+b) 	imes (a 	imes b) = a 	imes (a 	imes b) + b 	imes (a 	imes b)$. Using the vector triple product formula $A 	imes (B 	imes C) = (A \cdot C)B - (A \cdot B)C$: $a 	imes (a 	imes b) = (a \cdot b)a - (a \cdot a)b = (a \cdot b)a - b$. $b 	imes (a 	imes b) = (b \cdot b)a - (b \cdot a)b = a - (a \cdot b)b$. Adding these results: $(a \cdot b)a - b + a - (a \cdot b)b = a(a \cdot b + 1) - b(1 + a \cdot b) = (a - b)(a \cdot b + 1)$. Wait,re-evaluating the expansion: $a 	imes (a 	imes b) + b 	imes (a 	imes b) = (a \cdot b)a - (a \cdot a)b + (b \cdot b)a - (b \cdot a)b = (a \cdot b)a - b + a - (a \cdot b)b = a(1 + a \cdot b) - b(1 + a \cdot b) = (a - b)(1 + a \cdot b)$. Thus,the vector $(a+b) 	imes (a 	imes b)$ is parallel to the vector $(a - b)$.

If $a$ and $b$ are unit vectors,then the vector $(a+b) \times (a \times b)$ is parallel to the vector

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