Given the points A(3,3,-2) , B(0,-3,4) , C(0,-3,0)…

Given the points $A(3,3,-2)$, $B(0,-3,4)$, $C(0,-3,0)$, $D(0,2,-4)$, find $\mathrm{Pr}_{\overrightarrow{CD}}\overrightarrow{AB}$, $\mathrm{Pr}_{\overrightarrow{AB}}\overrightarrow{CD}$ and the angle between $\overrightarrow{AB}$ and $\overrightarrow{CD}$.

Explanation of Method

Projections of AB onto CD and of CD onto AB.

Step 1: The vectors

\[ \overrightarrow{AB}=(-3,-6,6),\qquad |\overrightarrow{AB}|=\sqrt{81}=9, \]

\[ \overrightarrow{CD}=(0,5,-4),\qquad |\overrightarrow{CD}|=\sqrt{41}. \]

Step 2: Scalar product

\[ \overrightarrow{AB}\cdot\overrightarrow{CD}=0-30-24=-54. \]

Step 3: Projections

\[\mathrm{Pr}_{\overrightarrow{CD}}\overrightarrow{AB}=\frac{-54}{\sqrt{41}}\approx-8.43\]

\[\mathrm{Pr}_{\overrightarrow{AB}}\overrightarrow{CD}=\frac{-54}{9}=-6\]

Step 4: Angle

\[ \cos\theta=\frac{-54}{9\sqrt{41}}=-\frac{6}{\sqrt{41}}\approx-0.9370. \]

\[\theta=\arccos\left(-\frac{6}{\sqrt{41}}\right)\approx159.6^\circ\]

$A$ $B$ $C$ $D$ $\text{foot}$

$\mathrm{Pr}_{\overrightarrow{CD}}\overrightarrow{AB}\approx-8.43$
$\text{line through }A\parallel\overrightarrow{CD}$

Projection of $\overrightarrow{AB}$ onto the direction of $\overrightarrow{CD}$: drop $B$ onto the line through $A$ parallel to $CD$. The foot lies on the side opposite to $\overrightarrow{CD}$ (negative).

$C$ $D$ $A$ $B$ $\text{foot}$

$\mathrm{Pr}_{\overrightarrow{AB}}\overrightarrow{CD}=-6$
$\text{line through }C\parallel\overrightarrow{AB}$

Projection of $\overrightarrow{CD}$ onto the direction of $\overrightarrow{AB}$: drop $D$ onto the line through $C$ parallel to $AB$; signed length $-6$.

Result

\[\boxed{\,\mathrm{Pr}_{\overrightarrow{CD}}\overrightarrow{AB}=\frac{-54}{\sqrt{41}}\approx-8.43\,}\]

\[\boxed{\,\mathrm{Pr}_{\overrightarrow{AB}}\overrightarrow{CD}=\frac{-54}{9}=-6\,}\]

\[\boxed{\,\theta=\arccos\left(-\frac{6}{\sqrt{41}}\right)\approx159.6^\circ\,}\]

Remark

Correction. $\arccos(-6/\sqrt{41})=159.56^\circ$, not $159.4^\circ$.