For any non-zero vectors a, b, c,a cdot[(b+c) | vedclass

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(A) We have the expression $a \cdot[(b+c) 	imes(a+b+c)]$. Using the distributive property of the cross product,we expand the term inside the bracket:

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(A) We have the expression $a \cdot[(b+c) 	imes(a+b+c)]$. Using the distributive property of the cross product,we expand the term inside the bracket: $(b+c) 	imes(a+b+c) = (b 	imes a) + (b 	imes b) + (b 	imes c) + (c 	imes a) + (c 	imes b) + (c 	imes c)$. Since the cross product of any vector with itself is zero ($b 	imes b = 0$ and $c 	imes c = 0$) and $c 	imes b = -(b 	imes c)$,the expression simplifies to: $(b 	imes a) + (b 	imes c) + (c 	imes a) - (b 	imes c) = (b 	imes a) + (c 	imes a)$. Now,taking the dot product with $a$: $a \cdot [(b 	imes a) + (c 	imes a)] = a \cdot (b 	imes a) + a \cdot (c 	imes a)$. By the definition of the scalar triple product,$a \cdot (b 	imes a) = [a \ b \ a]$ and $a \cdot (c 	imes a) = [a \ c \ a]$. Since any scalar triple product with two identical vectors is zero,we have $[a \ b \ a] = 0$ and $[a \ c \ a] = 0$. Therefore,the final result is $0 + 0 = 0$.

For any non-zero vectors $a, b, c$,$a \cdot[(b+c) \times(a+b+c)] = \ldots .$

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