Question

The vectors $\overrightarrow{AB}=3\hat{i}-2\hat{j}+3\hat{k}$ and $\overrightarrow{BC}=-\hat{i}+2\hat{k}$ are the adjacent sides of a parallelogram $ABCD$ then the

• A. ${\cos }^{-1}\left(\sqrt{\frac{1}{85}}\right)$
• B. $\pi {\cos }^{-1}\left(\sqrt{\frac{49}{85}}\right)$
• C. ${\cos }^{-1}\left(\frac{1}{2\sqrt{2}}\right)$
• D. ${\cos }^{-1}\left(\sqrt{\frac{3}{10}}\right)$

Answer 1

Answer: (A)

${\cos }^{-1}\left(\sqrt{\frac{1}{85}}\right)$

Given,

$AB=3i-2j+3k$

$BC=-i+2k$

$|AB|=\sqrt{17}$

$|BC|=\sqrt{5}$

$|AB+BC{|}^2=|AB{|}^2+|BC{|}^2+2|AB||BC|\cos \theta$

$|2i-2j+4k{|}^2=17+5+2\sqrt{17}\sqrt{5}\cos \theta$

$24=22+2\sqrt{85}\cos \theta$

$\frac{1}{\sqrt{85}}=\cos \theta$

$\therefore \theta ={\cos }^{-1}\frac{1}{\sqrt{85}}$