Answer
$cos^{-1}(\frac{7}{ \sqrt {130}})\approx 52^\circ$
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 $|a|=\sqrt {(a_{1})^{2}+(a_{2})^{2}+(a_{3})^{2}}$
$a.b=4 \times 2+ -3\times (0)+ 1 \times -1=8-0-1=7$
$|a|=\sqrt {(4)^{2}+(-3)^{2}+(1)^{2}}=\sqrt {26}$
$|b|=\sqrt {(2)^{2}+(0)^{2}+(-1)^{2}}=\sqrt {5}$
Angle between two vectors is given by
$cos\theta=\frac{a.b}{|a||b|}$
$cos\theta=\frac{7}{ \sqrt {26}\times \sqrt {5}}=\frac{7}{ \sqrt {130}}$
$\theta=cos^{-1}(\frac{7}{ \sqrt {130}})\approx 52^\circ$
Hence, $cos^{-1}(\frac{7}{ \sqrt {130}})\approx 52^\circ$
