Question

If $AB=-i-2j-6k,BC=2i-j+k,AC=i-3j-5k$. 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

Answer: (B)

${\cos }^{-1}\left(\sqrt{\frac{6}{41}}\right)$

$\cos \angle B=|\frac{a_1{b}_1+a_2{b}_2+a_3{b}_3}{\sqrt{a_1^2+a_2^2+a_3^2}\sqrt{b_1^2+b_2^2+b_3^2}}|$

$=|\frac{-2+2-6}{\sqrt{1+4+36}\sqrt{4+1+1}}|$

$=|\frac{-6}{\sqrt{41}\sqrt{6}}|$

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

$\therefore \angle B={\cos }^{-1}\left(\sqrt{\frac{6}{41}}\right)$

Option B is correct