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{\bf Question}

Calculate  $\mathbf {u} + \mathbf {v}, \mathbf {u} - \mathbf {v},
|\mathbf {u}|, |\mathbf {v}|, \hat \mathbf {u}, \hat \mathbf {v},
\mathbf{u} \cdot \mathbf {v}$, the angle between  $\mathbf {u}$
and $\mathbf {v}$, the scalar projection of $\mathbf {u}$ in the
direction of $\mathbf {v}$, and the vector projection of $\mathbf
{u}$ along $\mathbf {v}$ for:

\begin{description}

\item[(a)]
$\mathbf {u=i-j,\ v=\textrm {2}i+j+\textrm {2}k};$

\item[(b)]
$\mathbf {u=i+\textrm{2}k,\ v=j+k};$

\item[(c)]
$\mathbf {u=\textrm{2}i+\textrm{4}j-\textrm{3}k,\ v=i+j+k}.$

\end{description}



{\bf Answer}

\begin{description}
\item[(a)]
$\displaystyle {\bf u}+{\bf v}=3{\bf i}+2{\bf k},\ {\bf u}-{\bf
v}=-{\bf i}-2{\bf j}-2{\bf k},\ |{\bf u}|=\sqrt{2},\ |{\bf v}|=3,$

$\displaystyle {\bf \hat u}=\frac{1}{\sqrt{2}}{\bf
i}-\frac{1}{\sqrt{2}}{\bf j},\ {\bf \hat v}=\frac{2}{3}{\bf
i}+\frac{1}{3}{\bf j}+\frac{2}{3}{\bf k},\ {\bf u} \cdot {\bf
v}=2-1+0=1,$

$\displaystyle \theta=\cos^{-1} \left (\frac{{\bf u}\cdot{\bf
v}}{|{\bf u}| |{\bf v}|} \right)=\cos^{-1} \left
(\frac{1}{3\sqrt{2}} \right)\approx 76.37^{\circ},\ {\bf u} \cdot
{\bf \hat v}=\frac{{\bf u}\cdot{\bf v}}{|{\bf v}|}=\frac{1}{3},$

$\displaystyle ({\bf u}\cdot{\bf \hat v}){\bf \hat
v}=\frac{2}{9}{\bf i}+\frac{1}{9}{\bf j}+\frac{2}{9}{\bf k}.$

\item[(b)]
$\displaystyle {\bf u}+{\bf v}={\bf i}+{\bf j}+3{\bf k},\ {\bf
u}-{\bf v}={\bf i}-{\bf j}+{\bf k},\ |{\bf u}|=\sqrt{5},\ |{\bf
v}|=\sqrt{2},$

$\displaystyle {\bf \hat u}=\frac{1}{\sqrt{5}}{\bf
i}+\frac{2}{\sqrt{5}}{\bf k},\ {\bf \hat v}=\frac{1}{\sqrt{2}}{\bf
j}+\frac{1}{\sqrt{2}}{\bf k},\ {\bf u} \cdot {\bf v}=0+0+2=2,$

$\displaystyle \theta=\cos^{-1} \left (\frac{2}{\sqrt{5}\times
\sqrt{2}} \right)\approx 50.77^{\circ},\ {\bf u} \cdot {\bf \hat
v}=\frac{{\bf u}\cdot{\bf v}}{|{\bf v}|}=\frac{2}{\sqrt
2}=\sqrt{2},$

$\displaystyle ({\bf u}\cdot{\bf \hat v}){\bf \hat v}={\bf j}+{\bf
k}.$

\item[(c)]
$\displaystyle {\bf u}+{\bf v}=3{\bf i}+5{\bf j}-2{\bf k},\ {\bf
u}-{\bf v}={\bf i}+3{\bf j}-4{\bf k},\ |{\bf u}|=\sqrt{29},\ |{\bf
v}|=\sqrt{3},$

$\displaystyle {\bf \hat u}=\frac{2}{\sqrt{29}}{\bf
i}+\frac{4}{\sqrt{29}}{\bf j}-\frac{3}{\sqrt{29}}{\bf k},\ {\bf
\hat v}=\frac{1}{\sqrt{3}}{\bf i}+\frac{1}{\sqrt{3}}{\bf
j}+\frac{1}{\sqrt{3}}{\bf k},$

$\displaystyle {\bf u} \cdot {\bf v}=2+4-3=3,\ \theta=\cos^{-1}
\left (\frac{3}{\sqrt{29}\times \sqrt{3}} \right)\approx
71.24^{\circ},$

$\displaystyle {\bf u} \cdot {\bf \hat v}=\frac{{\bf u}\cdot{\bf
v}}{|{\bf v}|}=\frac{3}{\sqrt 3}=\sqrt{3},\ ({\bf u}\cdot{\bf \hat
v}){\bf \hat v}={\bf i}+{\bf j}+{\bf k}.$
\end{description}


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