## 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 $a_{1}b_{1}+a_{2}b_{2}+a_{3}b_{3}$. Therefore, $<0,1,0>\cdot<7,41,-3>=$$0\times7+1\times4+0\times(-3)= 41$