If $\left(\mathrm{l}_1, \mathrm{~m}_1, \mathrm{n}_1\right),\left(\mathrm{l}_2, \mathrm{~m}_2,…

If $\left(\mathrm{l}_1, \mathrm{~m}_1, \mathrm{n}_1\right),\left(\mathrm{l}_2, \mathrm{~m}_2, \mathrm{n}_2\right)$ are the direction cosines of two lines, then $\left(1_1 m_2-1_2 m_1\right)^2+\left(m_1 n_2-m_2 n_1\right)^2+$ $\left(\mathrm{n}_1 \mathrm{l}_2-\mathrm{n}_2 \mathrm{l}_1\right)^2+\left(\mathrm{l}_1 \mathrm{l}_2+\mathrm{m}_1 \mathrm{~m}_2+\mathrm{n}_1 \mathrm{n}_2\right)^2=$
  1. $0$
  2. $1$
  3. $2$
  4. $4$

Solution

Given $\left(l_1, m_1, \mathrm{n}_1\right),\left(l_2, m_2, n_2\right)$ are d.cs of two lines. $\begin{aligned} & \text { So } l_1^2+m_1^2+n_1^2=1, l_2^2+m_2^2+n_2^2=1 \\ & \text { Now, }\left(l_1 m_2-l_2 m_1\right)^2+\left(m_1 n_2-m_2 n_1\right)^2+\left(n_1 l_2-n_2 l_1\right)^2+\left(l_1 l_2\right. \\ & \left.+m_1 m_2+n_1 n_2\right)^2 \\ & =l_1^2 m_2^2+l_2^2 m_1^2-2 l_1 l_2 m_1 m_2+m_1^2 n_2^2+m_2^2 n_1^2 \\ & -2 m_1 m_2 n_1 n_2+n_1^2 l_2^2+n_2^2 l_1^2-2 l_1 l_2 n_1 n_2+ \\ & \quad l_1^2 l_2^2+m_1^2 m_2^2+n_1^2 n_2^2 \\ & +2 l_1 l_2 m_1 m_2+2 m_1 m_2 n_1 n_2+2 l_1 l_2 n_1 n_2 \\ & =l_2^2\left(l_1^2+m_1^2+n_1^2\right)+m_2^2\left(m_1^2+n_1^2+l_1^2\right) \\ & +n_2^2\left(l_1^2+m_1^2+n_1^2\right)=l_2^2+m_2^2+n_2^2=1 \end{aligned}$

Asked in: AP EAMCET 2023 (16 May Shift 2)

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