Answer
The projection of $\overrightarrow {AC} $ along $\overrightarrow {AD} $ is $\left( {0,1, - 1} \right)$, which is $\overrightarrow {AD} $ itself.
Work Step by Step
From Figure 16, we have $A = \left( {0,0,1} \right)$, $C = \left( {1,1,0} \right)$, $D = \left( {0,1,0} \right)$.
Write ${\bf{u}} = \overrightarrow {AC} $ and ${\bf{v}} = \overrightarrow {AD} $. So,
${\bf{u}} = \overrightarrow {AC} = C - A = \left( {1,1,0} \right) - \left( {0,0,1} \right) = \left( {1,1, - 1} \right)$
${\bf{v}} = \overrightarrow {AD} = D - A = \left( {0,1,0} \right) - \left( {0,0,1} \right) = \left( {0,1, - 1} \right)$
By Eq. (4) of Theorem 3, the projection of ${\bf{u}} = \overrightarrow {AC} $ along ${\bf{v}} = \overrightarrow {AD} $ is the vector ${{\bf{u}}_{||{\bf{v}}}}$ given by
${{\bf{u}}_{||{\bf{v}}}} = \left( {\frac{{{\bf{u}}\cdot{\bf{v}}}}{{{\bf{v}}\cdot{\bf{v}}}}} \right){\bf{v}}$
${{\bf{u}}_{||{\bf{v}}}} = \left( {\frac{{\left( {1,1, - 1} \right)\cdot\left( {0,1, - 1} \right)}}{{\left( {0,1, - 1} \right)\cdot\left( {0,1, - 1} \right)}}} \right)\left( {0,1, - 1} \right)$
${{\bf{u}}_{||{\bf{v}}}} = \left( {\frac{2}{2}} \right)\left( {0,1, - 1} \right) = \left( {0,1, - 1} \right)$
But $\overrightarrow {AD} = \left( {0,1, - 1} \right)$. Thus, the projection of $\overrightarrow {AC} $ along $\overrightarrow {AD} $ is $\overrightarrow {AD} $ itself.
