If u, v and w are three non-coplanar vectors, | vedclass

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(B) We need to evaluate the scalar triple product $(u + v - w) \cdot [(u - v) 	imes (v - w)]$.First,expand the cross product: $(u - v) 	imes (v - w)

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$u \cdot (v 	imes w)$

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(B) We need to evaluate the scalar triple product $(u + v - w) \cdot [(u - v) 	imes (v - w)]$.First,expand the cross product: $(u - v) 	imes (v - w) = u 	imes v - u 	imes w - v 	imes v + v 	imes w$.Since $v 	imes v = 0$,this simplifies to $u 	imes v - u 	imes w + v 	imes w$.Now,compute the dot product with $(u + v - w)$:$(u + v - w) \cdot (u 	imes v - u 	imes w + v 	imes w) = u \cdot (u 	imes v) - u \cdot (u 	imes w) + u \cdot (v 	imes w) + v \cdot (u 	imes v) - v \cdot (u 	imes w) + v \cdot (v 	imes w) - w \cdot (u 	imes v) + w \cdot (u 	imes w) - w \cdot (v 	imes w)$.Using the property that the scalar triple product is zero if any two vectors are identical,we have:$u \cdot (u 	imes v) = 0, u \cdot (u 	imes w) = 0, v \cdot (u 	imes v) = 0, v \cdot (v 	imes w) = 0, w \cdot (u 	imes w) = 0, w \cdot (v 	imes w) = 0$.Thus,the expression simplifies to:$u \cdot (v 	imes w) - v \cdot (u 	imes w) - w \cdot (u 	imes v)$.Using the cyclic property of scalar triple products $[u, v, w] = [v, w, u] = [w, u, v]$ and $[v, u, w] = -[u, v, w]$:$= [u, v, w] - (-[u, v, w]) - [w, u, v] = [u, v, w] + [u, v, w] - [u, v, w] = [u, v, w] = u \cdot (v 	imes w)$.

If $u, v$ and $w$ are three non-coplanar vectors,then $(u + v - w) \cdot [(u - v) \times (v - w)]$ equals

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