Question

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 the lines $CA$ and $BD$ is:

• A. ${\cos }^{-1}\left(\frac{-55}{\sqrt{149}\sqrt{161}}\right)$
• B. ${\cos }^{-1}\left(\frac{65}{\sqrt{149}\sqrt{161}}\right)$
• C. ${\cos }^{-1}\left(\frac{15}{\sqrt{149}\sqrt{161}}\right)$
• D. ${\cos }^{-1}\left(\frac{3}{\sqrt{149}\sqrt{161}}\right)$

Answer 1

The correct option is A ${\cos }^{-1}\left(\frac{-55}{\sqrt{149}\sqrt{161}}\right)$

$E$ is the mid point of $CA$ and $BD$
$\therefore E=(-2,2,\frac{9}{2})$
Since it is also mid point of $BD$, we have $D=(-4,-2,9)$
$D.R^{\prime }s$ of $CA$ are $(12,2,1)$
$D.C^{\prime }s$ of $CA$ are $(\frac{12}{\sqrt{149}},\frac{2}{\sqrt{149}},\frac{1}{\sqrt{149}})=(l_1,m_1,n_1)$
$D.R^{\prime }s$ of $BD$ are $(-4,-8,9)$
$D.C^{\prime }s$ of $BD$ are $(\frac{-4}{\sqrt{161}},\frac{-8}{\sqrt{161}},\frac{9}{\sqrt{161}})=(l_2,m_2,n_2)$
Now, angle between the diagonals is $\cos \theta =l_1.l_2+m_1.m_2+n_1.n_2$
$\therefore \theta ={\cos }^{-1}\left(\frac{-55}{\sqrt{149}\sqrt{161}}\right)$