Answer
a) $\frac{4}{\sqrt{17}} \hat{\jmath}+-\frac{\hat{i}}{\sqrt{17}}$
b)
$\left.-\frac{2}{\sqrt{14}} \hat{\jmath}+\frac{\hat{k}}{\sqrt{14}} \quad c+\frac{3}{\sqrt{14}} \hat{\imath}\right)\left\langle\frac{2 \sqrt{2}}{3}, \frac{\sqrt{2}}{6},-\frac{\sqrt{2}}{6}\right\rangle$
Work Step by Step
The unitary vector in the direction of a vector $\vec{u}$ is
\[
\hat{u}=\vec{u} \frac{1}{\|\vec{u}\|}
\]
For $\vec{u}=4 \hat{\jmath}-\hat{\imath}$ we get that
\[
\hat{u}=\frac{1}{\|4\hat{\jmath}- \hat{\imath}\|} \quad(4\hat{\jmath}- \hat{\imath})=\frac{1}{\sqrt{4^{2}+(-1)^{2}}}(-\hat{\imath}+4 \hat{\jmath})=\frac{4 \hat{\jmath}}{\sqrt{17}}-\frac{ \hat{\imath}}{\sqrt{17}}
\]
b) The unitary vector in the opposite direction of $\vec{u}=-4 \hat{\jmath}+2 \hat{k}+6 \hat{\imath}$ is
\[
-\hat{u}=-\vec{u} \frac{1}{\|\vec{u}\|} =-\frac{1}{\sqrt{(-4)^{2}+2^{2}+6^{2}}},(6 \hat{\imath}-4 \hat{\jmath}+2 \hat{k})=-\frac{2}{\sqrt{14}} \hat{\jmath}+\frac{\hat{k}}{\sqrt{14}}
+\frac{3}{\sqrt{14}} \hat{\imath}\]
c) The vector from $A(-1,0,2)$ to the point $B(3,1,1)$ is
\[
\overrightarrow{A B}=\vec{v}=\langle -(-1)+3, 1-0,-2+1\rangle=\langle 4,1,-1\rangle
\]
and then the unitary vector is
\[
\hat{v}=\frac{\vec{v}}{\|\vec{v}\|}=\langle 4,1,-1\rangle \frac{1}{\sqrt{1^{2}+(-1)^{2}+4^{2}}}=\langle 4,1,-1\rangle \frac{1}{3 \sqrt{2}}
\]
