Find the triple scalar product. $\mathbf { u } = ( 2,3,3 ) , \mathbf { v } = ( 1,2,0 ) , \mathbf { w } = ( 0,0,4 )$
A) $\mathbf { u } \cdot ( \mathbf { v } \times \mathbf { w } ) = 4$
B) $\mathbf { u } \cdot ( \mathbf { v } \times \mathbf { w } ) = 5$
C) $\mathbf { u } \cdot ( \mathbf { v } \times \mathbf { w } ) = 6$
D) $\mathbf { u } \cdot ( \mathbf { v } \times \mathbf { w } ) = 3$
E) $\mathbf { u } \cdot ( \mathbf { v } \times \mathbf { w } ) = 2$

Question

Find the triple scalar product.  $\mathbf { u } = ( 2,3,3 ) , \mathbf { v } = ( 1,2,0 ) , \mathbf { w } = ( 0,0,4 )$ 
A) $\mathbf { u } \cdot ( \mathbf { v } \times \mathbf { w } ) = 4$
B) $\mathbf { u } \cdot ( \mathbf { v } \times \mathbf { w } ) = 5$
C) $\mathbf { u } \cdot ( \mathbf { v } \times \mathbf { w } ) = 6$
D) $\mathbf { u } \cdot ( \mathbf { v } \times \mathbf { w } ) = 3$
E) $\mathbf { u } \cdot ( \mathbf { v } \times \mathbf { w } ) = 2$

Quizplus · Accepted Answer

The triple scalar product $\mathbf { u } \cdot ( \mathbf { v } \times \mathbf { w } )$ can be calculated by first finding the cross product $\mathbf { v } \times \mathbf { w }$, and then taking the dot product of the result with $\mathbf { u }$. The cross product $\mathbf { v } \times \mathbf { w }$ is calculated as follows:$$\mathbf { v } \times \mathbf { w } = \left| \begin{array}{ccc}\mathbf{i} & \mathbf{j} & \mathbf{k} \1 & 2 & 0 \0 & 0 & 4 \\end{array} \right| = (8\mathbf{j}) - (0\mathbf{k}) - (0\mathbf{i}) = (0, 8, 0)$$Then, the dot product $\mathbf { u } \cdot ( \mathbf { v } \times \mathbf { w } )$ is:$$\mathbf { u } \cdot (0, 8, 0) = (2, 3, 3) \cdot (0, 8, 0) = 0*2 + 8*3 + 0*3 = 24$$It appears there was a mistake in my initial calculation as none of the provided options match the result of 24. Let's correct the process:First, calculate the cross product $\mathbf { v } \times \mathbf { w }$:$$\mathbf { v } \times \mathbf { w } = \left| \begin{array}{ccc}\mathbf{i} & \mathbf{j} & \mathbf{k} \1 & 2 & 0 \0 & 0 & 4 \\end{array} \right| = (8\mathbf{i}) - (4\mathbf{j}) + (0\mathbf{k}) = (8, -4, 0)$$Then, calculate the dot product $\mathbf { u } \cdot ( \mathbf { v } \times \mathbf { w } )$:$$\mathbf { u } \cdot (8, -4, 0) = (2, 3, 3) \cdot (8, -4, 0) = 16 - 12 + 0 = 4$$The correct calculation shows that the triple scalar product is indeed 4, which matches option A. My initial explanation contained a calculation error, and the correct answer is based on the corrected calculation.

Find the Triple Scalar Product u=(2,3,3),v=(1,2,0),w=(0,0,4)\mathbf { u } = ( 2,3,3 ) , \mathbf { v } = ( 1,2,0 ) , \mathbf { w } = ( 0,0,4 )u=(2,3,3),v=(1,2,0),w=(0,0,4)

Find the Triple Scalar Product $\mathbf { u } = ( 2,3,3 ) , \mathbf { v } = ( 1,2,0 ) , \mathbf { w } = ( 0,0,4 )$