Question

If three consecuting 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 anlge between $\overline{AC}$ and $\overline{BD}$ 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 ${\cos }^{-1}\left[\frac{55}{\sqrt{149}\sqrt{161}}\right]$

$\frac{a+0}{2}=\frac{4-8}{2}\Rightarrow a=-4$

$\frac{b+6}{2}=\frac{3+1}{2}\Rightarrow b=-2$

$\frac{c+0}{2}=\frac{5+4}{2}\Rightarrow c=9$

$\overrightarrow{AC}=12\hat{i}-2\hat{j}-\hat{k}$ $\overrightarrow{AD}=-4\hat{i}+8\hat{j}+9\hat{b}$

$\overrightarrow{AC}=\overrightarrow{BD}=|AC||BD|\cos (\widehat{AC}.BD)$

$\cos \left(\widehat{AC}BD\right)=\frac{\overrightarrow{AC}.\overrightarrow{BD}}{|AC||BD|}=\frac{48+16-9}{\sqrt{149}\sqrt{161}}=\frac{55}{\sqrt{149}\sqrt{169}}$

$\widehat{AC}.BC={\cos }^{-1}\left(\frac{55}{\sqrt{149}\sqrt{169}}\right)$