If three consecutive vertices of a parallelogram are A(4,3,5), B(0,6,0), C(–8,1,4) and D is the fourth vertex then the angle between $\overleftrightarrow{\mathrm{AC}}\text{ and }\overleftrightarrow{\mathrm{BD}}$ is

In the figure E is mid point of AC and BD
$\therefore \mathrm{E}=\left(-2,2,\frac{9}{2}\right)$
Since it is also midpoint BD, we have D


4,
2,9
D.r’s of AC are (12, 2, 1)
D.c’s of AC are

$\left(\frac{12}{\sqrt{149}},\frac{2}{\sqrt{149}},\frac{1}{\sqrt{149}}\right)$
D.c’s of BD are $\left(\frac{-4}{\sqrt{161}},\frac{-8}{\sqrt{161}},\frac{9}{\sqrt{161}}\right)$

Angle between diagonals $\theta ={\mathrm{Cos}}^{-1}\left(\frac{55}{\sqrt{149}\sqrt{161}}\right)$