Answer
$cos^{-1}(\frac{10}{{3\times 5}})=cos^{-1}(\frac{10}{{15}})\approx 48^\circ$
Work Step by Step
The dot product of $a=a_{1}i+a_{2}j+a_{3}k$ and $b=b_{1}i+b_{2}j+b_{3}k$ is defined as $a.b=a_{1}\times b_{1}+a_{2}\times b_{2}+a_{3}\times b_{3}$
The magnitude of a vector is given as $|a|=\sqrt {(a_{1})^{2}+(a_{2})^{2}+(a_{3})^{2}}$
$a.b=1 \times 4+ 2\times 0+ -2 \times -3=4+0+6=10$
$|a|=\sqrt {(1)^{2}+(2)^{2}+(-2)^{2}}=\sqrt {9}=3$
$|b|=\sqrt {(4)^{2}+(0)^{2}+(-3)^{2}}=\sqrt {25}=5$
Angle between two vectors is given by
$cos\theta=\frac{a.b}{|a||b|}$
$cos\theta=\frac{10}{3\times 5}$
$\theta=cos^{-1}(\frac{10}{{3\times 5}})=cos^{-1}(\frac{10}{{15}})\approx 48^\circ$
Hence, $cos^{-1}(\frac{10}{{3\times 5}})=cos^{-1}(\frac{10}{{15}})\approx 48^\circ$
