If (l_1, m_1, n_1) and (l_2, m_2, n_2) are th | vedclass

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(B) Given that $(l_1, m_1, n_1)$ and $(l_2, m_2, n_2)$ are the direction cosines of two lines. Since they are direction cosines,we have $l_1^2 + m_1^2

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(B) Given that $(l_1, m_1, n_1)$ and $(l_2, m_2, n_2)$ are the direction cosines of two lines. Since they are direction cosines,we have $l_1^2 + m_1^2 + n_1^2 = 1$ and $l_2^2 + m_2^2 + n_2^2 = 1$. Let $	heta$ be the angle between the two lines. Then $\cos 	heta = l_1 l_2 + m_1 m_2 + n_1 n_2$. The expression is $(l_1 m_2 - l_2 m_1)^2 + (m_1 n_2 - m_2 n_1)^2 + (n_1 l_2 - n_2 l_1)^2 + (l_1 l_2 + m_1 m_2 + n_1 n_2)^2$. By Lagrange's Identity,$(l_1 m_2 - l_2 m_1)^2 + (m_1 n_2 - m_2 n_1)^2 + (n_1 l_2 - n_2 l_1)^2 = (l_1^2 + m_1^2 + n_1^2)(l_2^2 + m_2^2 + n_2^2) - (l_1 l_2 + m_1 m_2 + n_1 n_2)^2$. Substituting the values,we get $(1)(1) - \cos^2 	heta = \sin^2 	heta$. Thus,the total expression becomes $\sin^2 	heta + \cos^2 	heta = 1$.

If $(l_1, m_1, n_1)$ and $(l_2, m_2, n_2)$ are the direction cosines of two lines,then $(l_1 m_2 - l_2 m_1)^2 + (m_1 n_2 - m_2 n_1)^2 + (n_1 l_2 - n_2 l_1)^2 + (l_1 l_2 + m_1 m_2 + n_1 n_2)^2 =$

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