For vectors $\mathbf{A}=\langle 2,-1,3\rangle$ and $\mathbf{B}=\langle 4,0,-2\rangle$, the dot product $\mathbf{A}\cdot\mathbf{B}$ is
A$2$
B$4$
C$8$
D$-2$
Answer & Solution
Correct answer: A. $2$
Using $\mathbf{A}\cdot\mathbf{B}=A_1B_1+A_2B_2+A_3B_3$, we get $2\cdot 4+(-1)\cdot 0+3\cdot(-2)=8+0-6=2$.
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