Question

If $\overline{AB}=-\overline{i}-2\overline{j}-6\overline{k},\overline{BC}=2\overline{i}-\overline{j}+\overline{k},\overline{AC}=\overline{i}-3\overline{j}-5\overline{k}$. Then $\angle B=?$

• A. ${\cos }^{-1}\left(\sqrt{\frac{40}{41}}\right)$
• B. ${\cos }^1\left(\sqrt{\frac{6}{41}}\right)$
• C. ${\cos }^{-1}\left(\frac{6}{41}\right)$
• D. ${\cos }^{-1}\left(\frac{62}{63}\right)$

Answer 1

The correct option is B ${\cos }^1\left(\sqrt{\frac{6}{41}}\right)$

$AB.BC=(-\hat{i}-2\hat{j}-6\hat{k}).(2\hat{i}-\hat{j}+\hat{k})$

$|AB|.|BC|cosB=6$

$\sqrt{41}\sqrt{6}cosB=6$

$B={\cos }^{-1}\sqrt{\frac{6}{41}}$