## Calculus (3rd Edition)

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$$(i+j+k)\cdot (3i+2j-5k)=3i\cdot i+2i\cdot j-5i\cdot k+3j\cdot i+2j\cdot j-5j\cdot k\\ +3k\cdot i+2k\cdot j-5k\cdot k=3+2-5\\ =0.$$ Recall that: $i\cdot j=j\cdot i=k\cdot j=j\cdot k=i\cdot k=k\cdot i=0$ These dot products are zero because the vectors are perpendicular ($\theta=90^{\circ}$ and $\cos \theta=0$). Recall that: $i\cdot i=j\cdot j=k\cdot k=1$ These dot products are 1 because the vectors are parallel ($\theta=0^{\circ}$ and $\cos \theta=1$) and their length is 1 (unit vectors).