Answer
$cos^{-1}(\frac{4}{9\times \sqrt {20}})\approx 84.3^\circ$
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 $|a|=\sqrt {(a_{1})^{2}+(a_{2})^{2}+(a_{3})^{2}}$
$a.b=8 \times 0+ -1\times (4)+ 4 \times 2=0-4+8=4$
$|a|=\sqrt {(8)^{2}+(-1)^{2}+(4)^{2}}=\sqrt {81}=9$
$|b|=\sqrt {(0)^{2}+(4)^{2}+(2)^{2}}=\sqrt {20}$
Angle between two vectors is given by
$cos\theta=\frac{a.b}{|a||b|}$
$cos\theta=\frac{4}{9\times \sqrt {20}}$
$\theta=cos^{-1}(\frac{4}{9\times \sqrt {20}})\approx 84.3^\circ$
Hence, $cos^{-1}(\frac{4}{9\times \sqrt {20}})\approx 84.3^\circ$
