Let A(1,-1,2), B(6,11,2), C(1,2,6) be three p | vedclass

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(B) The direction ratios of $AB$ are $a_1 = 6-1 = 5$,$b_1 = 11-(-1) = 12$,$c_1 = 2-2 = 0$. The magnitude of $AB$ is $\sqrt{5^2+12^2+0^2} = \sqrt{25+14

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$\frac{36}{65}$

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(B) The direction ratios of $AB$ are $a_1 = 6-1 = 5$,$b_1 = 11-(-1) = 12$,$c_1 = 2-2 = 0$. The magnitude of $AB$ is $\sqrt{5^2+12^2+0^2} = \sqrt{25+144} = \sqrt{169} = 13$. Thus,the direction cosines of $AB$ are $l_1 = \frac{5}{13}$,$m_1 = \frac{12}{13}$,$n_1 = 0$. The direction ratios of $AC$ are $a_2 = 1-1 = 0$,$b_2 = 2-(-1) = 3$,$c_2 = 6-2 = 4$. The magnitude of $AC$ is $\sqrt{0^2+3^2+4^2} = \sqrt{9+16} = \sqrt{25} = 5$. Thus,the direction cosines of $AC$ are $l_2 = 0$,$m_2 = \frac{3}{5}$,$n_2 = \frac{4}{5}$. The value of $|l_1 l_2 + m_1 m_2 + n_1 n_2|$ is the cosine of the angle between the lines $AB$ and $AC$. $|l_1 l_2 + m_1 m_2 + n_1 n_2| = |(\frac{5}{13} 	imes 0) + (\frac{12}{13} 	imes \frac{3}{5}) + (0 	imes \frac{4}{5})| = |0 + \frac{36}{65} + 0| = \frac{36}{65}$.

Let $A(1,-1,2), B(6,11,2), C(1,2,6)$ be three points. If $l_1, m_1, n_1$ are the direction cosines of $AB$ and $l_2, m_2, n_2$ are the direction cosines of $AC$,then $|l_1 l_2+m_1 m_2+n_1 n_2|=$

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