Question

lf $\theta$ is the angle between two lines whose d.cs are $l_1,m_1,n_1$ and $l_2,m_2,n_2$, then

$\frac{\Sigma (l_1+l_2{)}^2}{4{\cos }^2\left(\frac{\theta }{2}\right)}+\frac{\Sigma (l_I-l_2{)}^2}{4{\sin }^2\left(\frac{\theta }{2}\right)}=$

• A. $1$
• B. $0$
• C. $-1$
• D. $2$

Answer 1

The correct option is D $2$

Let $\overrightarrow{a_1}$ and $\overrightarrow{a_2}$ be unit vectors along the lines.

$\overrightarrow{a_1}=l_1\hat{i}+m_1\hat{j}+n_1\hat{k}$

$\overrightarrow{a_2}=l_2\hat{i}+m_2\hat{j}+n_2\hat{k}$

$\Rightarrow l_1^2+m_1^2+n_1^2=1$ and $l_2^2+m_2^2+n_2^2=1$

Consider the dot product of $\overrightarrow{a_1}$ and $\overrightarrow{a_2}$

$\overrightarrow{a_1}.\overrightarrow{a_2}=\left(l_1{l}_2\right)+\left(m_1{m}_2\right)+\left(n_1{n}_2\right)$

$\Rightarrow \cos \theta =\left(l_1{l}_2\right)+\left(m_1{m}_2\right)+\left(n_1{n}_2\right)$

$S=\frac{\Sigma (l_1+l_2{)}^2}{4{\cos }^2\left(\frac{\theta }{2}\right)}+\frac{\Sigma (l_I-l_2{)}^2}{4{\sin }^2\left(\frac{\theta }{2}\right)}$

$\Rightarrow S=\frac{2+2(l_1{l}_2+m_1{m}_2+n_1{n}_2)}{4{\cos }^2\frac{\theta }{2}}+\frac{2-2(l_1{l}_2+m_1{m}_2+n_1{n}_2)}{4{\sin }^2\frac{\theta }{2}}$

$\Rightarrow S=\frac{2+2\cos \theta }{4{\cos }^2\frac{\theta }{2}}+\frac{2-2\cos \theta }{4{\sin }^2\frac{\theta }{2}}$

Now, $1+\cos \theta =2{\cos }^2\frac{\theta }{2}$ and $1-\cos \theta =2{\sin }^2\frac{\theta }{2}$

$\Rightarrow S=\frac{2}{2}+\frac{2}{2}$

$\Rightarrow S=2$