a cdot [(b + c) times (a + b + c)] is equal t | vedclass

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(D) We are given the expression $a \cdot [(b + c) 	imes (a + b + c)]$. Using the distributive property of the cross product,we expand the term inside

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(D) We are given the expression $a \cdot [(b + c) 	imes (a + b + c)]$. Using the distributive property of the cross product,we expand the term inside the square brackets: $(b + c) 	imes (a + b + c) = (b 	imes a) + (b 	imes b) + (b 	imes c) + (c 	imes a) + (c 	imes b) + (c 	imes c)$. Since the cross product of any vector with itself is zero ($b 	imes b = 0$ and $c 	imes c = 0$),the expression simplifies to: $(b 	imes a) + (b 	imes c) + (c 	imes a) + (c 	imes b)$. Now,taking the dot product with $a$: $a \cdot [(b 	imes a) + (b 	imes c) + (c 	imes a) + (c 	imes b)] = a \cdot (b 	imes a) + a \cdot (b 	imes c) + a \cdot (c 	imes a) + a \cdot (c 	imes b)$. This can be written in terms of scalar triple products: $[a \, b \, a] + [a \, b \, c] + [a \, c \, a] + [a \, c \, b]$. In a scalar triple product,if any two vectors are identical,the value is $0$. Thus,$[a \, b \, a] = 0$ and $[a \, c \, a] = 0$. Also,$[a \, c \, b] = -[a \, b \, c]$. Substituting these values: $0 + [a \, b \, c] + 0 - [a \, b \, c] = 0$.

$a \cdot [(b + c) \times (a + b + c)]$ is equal to

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