Question

If $A=(2,1,3)$ , $B=(5,2,-1),C=(-3,3,-3)$ be the vertices of the parallelogram $ABCD$. Then the angle between its diagonals is

• A. ${\cos }^{-1}\left(\frac{13}{25\sqrt{13}}\right)$
• B. ${\cos }^{-1}\left(\frac{23}{25\sqrt{13}}\right)$
• C. ${\cos }^{-1}\left(\frac{33}{25\sqrt{13}}\right)$
• D. ${\cos }^{-1}\left(\frac{43}{25\sqrt{13}}\right)$

Answer 1

The correct option is D ${\cos }^{-1}\left(\frac{43}{25\sqrt{13}}\right)$

Let ${}^{\prime }{h}^{\prime }$ be mid point
So $G=(-\frac{1}{2},2,0)----1$ (from $A$ and $C$)

$G=(\frac{5+a}{2},\frac{2+b}{2},\frac{c-1}{2})-----2$ (from $B$ and $D$)

Compare 1 & 2

$a=-6,b=2c=1$

So $\overrightarrow{BD}=(-11,0,+2)$

$\overrightarrow{AC}=(-5,2,-6)$

$\overrightarrow{AC}.\overrightarrow{BD}=|\overrightarrow{AC}|\overrightarrow{BD}|\cos \theta$

$\frac{55-12}{25\sqrt{13}}=\cos \theta$

$Q={\cos }^{-1}\left(\frac{43}{25\sqrt{13}}\right)$