Question

The direction`cosines of a line equally inclined to three mutually perpendicular lines having D.C.'s as ${\ell }_1{m}_1{n}_1:{\ell }_2{m}_2{n}_2:{\ell }_3{m}_3{n}_3$ are

• A. $l_1+l_2+l_3,m_1+m_2+m_3,n_1+n_2+n_3$
• B. $(\pm \frac{1}{\sqrt{3}},\pm \frac{1}{\sqrt{3}},\pm \frac{1}{\sqrt{3}})$
  
• C. $(\pm \frac{1}{\sqrt{2}},\pm \frac{1}{\sqrt{3}},\pm \frac{1}{\sqrt{4}})$
  
• D. none of these

Answer 1

The correct option is B $(\pm \frac{1}{\sqrt{3}},\pm \frac{1}{\sqrt{3}},\pm \frac{1}{\sqrt{3}})$

We know that, ${\cos }^2\alpha +{\cos }^2\beta +{\cos }^2\gamma =1$

But given that,

$\alpha$ =$\beta$ = $\gamma$

${\cos }^2\alpha +{\cos }^2\alpha +{\cos }^2\alpha =1$

$3{\cos }^2\alpha =1$

$\cos \alpha =\pm \frac{1}{\sqrt{3}}$

Hence, direction cosine are $(\pm \frac{1}{\sqrt{3}},\pm \frac{1}{\sqrt{3}},\pm \frac{1}{\sqrt{3}})$