Question

Find the angle between the lines parallel to the following pairs of vectors $6i+2j+3k,3i+12j+4k$

• A. ${\cos }^{-1}\left(\frac{54}{91}\right)$
• B. ${\sin }^{-1}\left(\frac{54}{91}\right)$
• C. ${\cos }^{-1}\left(\frac{51}{91}\right)$
• D. ${\sin }^{-1}\left(\frac{51}{91}\right)$

Answer 1

The correct option is A ${\cos }^{-1}\left(\frac{54}{91}\right)$

$\cos \theta =\left|\frac{a_1{a}_2+b_1{b}_2+c_1{c}_2}{\sqrt{{\left(a_1\right)}^2+{\left(b_1\right)}^2+{\left(c_1\right)}^2}\sqrt{{\left(a_2\right)}^2+{\left(b_2\right)}^2+{\left(c_2\right)}^2}}\right|$

$=\left|\frac{18+24+12}{\sqrt{36+4+9}\sqrt{9+144+16}}\right|=\left|\frac{54}{\sqrt{49}\sqrt{169}}\right|=\frac{54}{91}$

$\therefore \theta ={\cos }^{-1}\frac{54}{91}$