The vector c cdot (b+c) times (a+b+c) is equa | vedclass

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(A) We need to evaluate the expression $c \cdot ((b+c) 	imes (a+b+c))$.Using the distributive property of the cross product,we have:$(b+c) 	imes (a+

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$c \cdot (b 	imes a)$

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(A) We need to evaluate the expression $c \cdot ((b+c) 	imes (a+b+c))$.Using the distributive property of the cross product,we have:$(b+c) 	imes (a+b+c) = b 	imes a + b 	imes b + b 	imes c + c 	imes a + c 	imes b + c 	imes c$.Since the cross product of any vector with itself is zero ($b 	imes b = 0$ and $c 	imes c = 0$),the expression simplifies to:$(b+c) 	imes (a+b+c) = b 	imes a + b 	imes c + c 	imes a + c 	imes b$.Now,taking the dot product with $c$:$c \cdot (b 	imes a + b 	imes c + c 	imes a + c 	imes b) = c \cdot (b 	imes a) + c \cdot (b 	imes c) + c \cdot (c 	imes a) + c \cdot (c 	imes b)$.Using the scalar triple product property $[x, y, z] = x \cdot (y 	imes z)$,we get:$= [c, b, a] + [c, b, c] + [c, c, a] + [c, c, b]$.Since any scalar triple product with two identical vectors is zero,we have:$[c, b, c] = 0$,$[c, c, a] = 0$,and $[c, c, b] = 0$.Thus,the expression simplifies to $[c, b, a] = c \cdot (b 	imes a)$.Since $c \cdot (b 	imes a) = -c \cdot (a 	imes b)$,and looking at the options,the correct form is $c \cdot (b 	imes a)$.

The vector $c \cdot (b+c) \times (a+b+c)$ is equal to

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