Answer
$cos^{-1}(\frac{-10}{ \sqrt {18}\times \sqrt {8}})\approx 146.4^\circ$
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 $|a|=\sqrt {(a_{1})^{2}+(a_{2})^{2}+(a_{3})^{2}}$
$a.b=0 \times 1+ 2\times (-4)+ -2 \times 1=0-8-2=-10$
$|a|=\sqrt {(1)^{2}+(-4)^{2}+(1)^{2}}=\sqrt {18}$
$|b|=\sqrt {(0)^{2}+(2)^{2}+(-2)^{2}}=\sqrt {8}$
Angle between two vectors is given by
$cos\theta=\frac{a.b}{|a||b|}$
$cos\theta=\frac{-10}{ \sqrt {18}\times \sqrt {8}}$
$\theta=cos^{-1}(\frac{-10}{ \sqrt {18}\times \sqrt {8}})\approx 146.4^\circ$
Hence, $cos^{-1}(\frac{-10}{ \sqrt {18}\times \sqrt {8}})\approx 146.4^\circ$