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प्रश्न
Find the angle between the lines whose direction cosines are given by the equations
l + 2m + 3n = 0 and 3lm − 4ln + mn = 0
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उत्तर
` \text{ Given } : `
\[l + 2m + 3n = 0 . . . (1)\]
\[3lm - 4\ln + mn = 0 . . . (2)\]
\[\text { From } \left( 1 \right), \text { we get } \]
\[l = - 2m - 3n\]
\[\text { Substituting } l = - 2m - 3n \text { in } \left( 2 \right), \text { we get } \]
\[3\left( - 2m - 3n \right)m - 4\left( - 2m - 3n \right)n + mn = 0\]
\[ \Rightarrow - 6 m^2 - 9mn + 8mn + 12 n^2 + mn = 0\]
\[ \Rightarrow 12 n^2 - 6 m^2 = 0 \]
\[ \Rightarrow m^2 = 2 n^2 \]
\[ \Rightarrow m = \sqrt{2}n, - \sqrt{2} n\]
\[\text{ If } m = \sqrt{2}n,\text { then by substituting } m = \sqrt{2}n \text { in } \left( 1 \right), \text { we get } l = n\left( - 2\sqrt{2} - 3 \right) . \]
\[\text { If } m = - \sqrt{2} n, \text { then by substituting }m = - \sqrt{2} n \text { in } \left( 1 \right), \text { we get } l = n\left( 2\sqrt{2} - 3 \right) . \]
\[\text { Thus, the direction ratios of the two lines are proportional to } n\left( - 2\sqrt{2} - 3 \right), \sqrt{2}n, n \text { and } n\left( 2\sqrt{2} - 3 \right), - \sqrt{2} n, n \text { or } \left( - 2\sqrt{2} - 3 \right), \sqrt{2} , 1 \text{ and } \left( - 2\sqrt{2} - 3 \right), - \sqrt{2}, 1 . \]
\[\text { Vectors parallel to these lines are } \]
\[ \vec{a} = \left( - 2\sqrt{2} - 3 \right) \hat{i} + \sqrt{2} \hat{j} + \hat{k} \]
\[ \vec{b} = \left( 2\sqrt{2} - 3 \right) \hat{i} - \sqrt{2} \hat{ j} + \hat{k} \]
\[\text{ If } \theta \text{ is the angle between the lines, then } \theta \text{ is also the angle between } \vec{a} \text{ and } \vec{b} . \]
\[\text{ Now}, \]
\[\cos \theta = \frac{\vec{a} . \vec{b}}{\left| \vec{a} \right| \left| \vec{b} \right|}\]
\[ = \frac{\left[ \left( - 2\sqrt{2} - 3 \right) \hat{i} + \sqrt{2} \hat{j} + \hat{k} \right] . \left[ \left( 2\sqrt{2} - 3 \right) \hat{i} - \sqrt{2} \hat{j} + \hat{k} \right]}{\sqrt{8 + 9 + 12\sqrt{2} + 2 + 1} \sqrt{8 + 9 - 12\sqrt{2} + 2 + 1}} \]
\[ = \frac{- \left( 8 - 9 \right) - 2 + 1}{\sqrt{20 + 12\sqrt{2}} \sqrt{20 - 12\sqrt{2}}} \]
\[ = \frac{0}{\sqrt{20 + 12\sqrt{2}} \sqrt{20 - 12\sqrt{2}}}\]
\[ = 0\]
\[ \Rightarrow \theta = \frac{\pi}{2}\]
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