AP DSC School Assistant Math Model Test 2

Concept:The scalar triple product $\vec{u}\cdot(\vec{v}\times\vec{w})$ is evaluated by expanding the vectors and using properties: cross product of a vector with itself is zero, and the triple product changes sign when two vectors are swapped. The given expression simplifies to zero because the three difference vectors are coplanar (their sum is zero).Explanation:Let $\vec{u} = \vec{a} - \vec{b}$, $\vec{v} = \vec{b} - \vec{c}$, $\vec{w} = \vec{c} - \vec{a}$. We need $(\vec{a}-\vec{b})\cdot[(\vec{b}-\vec{c})\times(\vec{c}-\vec{a})]$.First expand the cross product:$(\vec{b}-\vec{c})\times(\vec{c}-\vec{a}) = \vec{b}\times\vec{c} - \vec{b}\times\vec{a} - \vec{c}\times\vec{c} + \vec{c}\times\vec{a}$.Since $\vec{c}\times\vec{c}=0$, this becomes $\vec{b}\times\vec{c} - \vec{b}\times\vec{a} + \vec{c}\times\vec{a}$.Now take the dot product with $(\vec{a}-\vec{b})$:$(\vec{a}-\vec{b})\cdot[\vec{b}\times\vec{c} - \vec{b}\times\vec{a} + \vec{c}\times\vec{a}] = \vec{a}\cdot(\vec{b}\times\vec{c}) - \vec{a}\cdot(\vec{b}\times\vec{a}) + \vec{a}\cdot(\vec{c}\times\vec{a}) - \vec{b}\cdot(\vec{b}\times\vec{c}) + \vec{b}\cdot(\vec{b}\times\vec{a}) - \vec{b}\cdot(\vec{c}\times\vec{a})$.Any scalar triple product with two identical vectors is zero: $\vec{a}\cdot(\vec{b}\times\vec{a}) = 0$, $\vec{a}\cdot(\vec{c}\times\vec{a}) = 0$, $\vec{b}\cdot(\vec{b}\times\vec{c}) = 0$, $\vec{b}\cdot(\vec{b}\times\vec{a}) = 0$.We are left with $\vec{a}\cdot(\vec{b}\times\vec{c}) - \vec{b}\cdot(\vec{c}\times\vec{a})$.Now $\vec{b}\cdot(\vec{c}\times\vec{a}) = [\vec{b},\vec{c},\vec{a}]$ and by cyclic permutation $[\vec{b},\vec{c},\vec{a}] = [\vec{a},\vec{b},\vec{c}]$. Hence the expression becomes $[\vec{a},\vec{b},\vec{c}] - [\vec{a},\vec{b},\vec{c}] = 0$.Answer:The value is $0$, which corresponds to option A.