Answer
θ=48 °11'
Work Step by Step
u·v=1·4+2·0-2·(-3)=10
|u|=$\sqrt (1^2+2^2+(-2)^2$=3
|v|=$\sqrt (4^2+0+(-3)^2$=5
cos θ=$\frac{10}{3·5}$=$\frac{2}{3}$
θ=48 °11'
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