Answer
a) $=(\frac{3}{\sqrt {38}},\frac{2}{\sqrt {38}}, -\frac{5}{\sqrt {38}})$
b) $=(-\frac{3}{\sqrt {38}},-\frac{2}{\sqrt {38}}, \frac{5}{\sqrt {38}})$
Work Step by Step
a) The magnitude of this vector is given by $\sqrt {(3)^{2} +(2)^{2}+(-5)^{2}}=\sqrt {38}$.
Therefore a unit vector in direction $u$ is given by:
$\frac{1}{\sqrt {38}}(3,2,-5)$
$=(\frac{3}{\sqrt {38}},\frac{2}{\sqrt {38}}, -\frac{5}{\sqrt {38}})$
b) a unit vector in the opposite direction is given by multiplying the unit vector found in part a) by $(-1)$
This is equal to $(-1)\times(\frac{3}{\sqrt {38}},\frac{2}{\sqrt {38}}, -\frac{5}{\sqrt {38}})$
$=(-\frac{3}{\sqrt {38}},-\frac{2}{\sqrt {38}}, \frac{5}{\sqrt {38}})$