The position vectors of A,B,C,D are vec r_1=\5,2,-1\ , vec…

The position vectors of $A,B,C,D$ are $\vec r_1=\{5,2,-1\}$, $\vec r_2=\{1,-3,4\}$, $\vec r_3=\{-2,1,3\}$, $\vec r_4=\{2,6,-2\}$. $1^\circ$ Show that $ABCD$ is a parallelogram. $2^\circ$ Find the angle between its diagonals.

Explanation of Method

ABCD is a parallelogram; the angle between its diagonals.

Step 1: Parallelogram

\[ \overrightarrow{AB}=\vec r_2-\vec r_1=\{-4,-5,5\},\qquad \overrightarrow{DC}=\vec r_3-\vec r_4=\{-4,-5,5\}, \]

\[ \overrightarrow{BC}=\vec r_3-\vec r_2=\{-3,4,-1\},\qquad \overrightarrow{AD}=\vec r_4-\vec r_1=\{-3,4,-1\}. \]

Since $\overrightarrow{AB}=\overrightarrow{DC}$ and $\overrightarrow{AB}\not\parallel\overrightarrow{BC}$ (the points are not collinear), $ABCD$ is a parallelogram. The sides are $|AB|=\sqrt{66}$ and $|BC|=\sqrt{26}$, so it is not a rhombus.

Step 2: Diagonals

\[ \overrightarrow{AC}=\{-7,-1,4\},\quad |\overrightarrow{AC}|=\sqrt{66};\qquad \overrightarrow{BD}=\{1,9,-6\},\quad |\overrightarrow{BD}|=\sqrt{118}. \]

\[ \overrightarrow{AC}\cdot\overrightarrow{BD}=-7-9-24=-40. \]

Step 3: Angle

\[ \cos\angle(\overrightarrow{AC},\overrightarrow{BD})=\frac{-40}{\sqrt{66}\sqrt{118}}=-\frac{20}{\sqrt{1947}}\approx-0.4533,\qquad \angle\approx116.95^\circ. \]

The lines of the diagonals make the acute angle: $\varphi=\arccos\frac{20}{\sqrt{1947}}\approx63.05^\circ$.

$M$ $A(5,2,-1)$ $B(1,-3,4)$ $C(-2,1,3)$ $D(2,6,-2)$ $O$

$\overrightarrow{AC}=\{-7,-1,4\}$
$\overrightarrow{BD}=\{1,9,-6\}$

The diagonals meet at $M\left(\tfrac32,\tfrac32,1\right)$, the common midpoint. The vectors make $116.95^\circ$, so the lines make $63.05^\circ$.

Result

\[\boxed{\,\varphi=\arccos\frac{20}{\sqrt{1947}}\approx63.05^\circ\,}\]