## Calculus (3rd Edition)

We know that the scalar product of $a_{1}\hat{i}+a_{2}\hat{j}+a_{3}\hat{k}$ and $b_{1}\hat{i}+b_{2}\hat{j}+ b_{3}\hat{k}$ is equal to $a_{1}b_{1}+a_{2}b_{2}+a_{3}b_{3}$. Then, $<3,-2,2>\cdot<1,0,1>=$$3\times1+(-2\times0)+2\times1=5$