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| If the direction cosines of two lines are given by the equations 3l + m + 5n = 0 and 6mn -2nl + 5lm = 0 , Then what is the angle between the lines? |
\(\cos^{-1}\left(\frac{-1}{3}\right)\) \(\cos^{-1}\left(\frac{1}{6}\right)\) \(\cos^{-1}\left(\frac{-1}{6}\right)\) None of the above
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| \(\cos^{-1}\left(\frac{1}{6}\right)\) |
| \(\begin{aligned}\text{Given, }3l+m+5n&=0\ldots (1),\\ 6mn-2nl+5lm&=0\ldots (2)\\ \text{From (1), }m=-(3l+5n)&\\ \text{From (2), }6(3l+5n)n-2nl+5l(-3l-5n)&=0\\ (l+2n)(l+n)&=0\\ l+2n=0\ldots (3) \text{or }l+n&=0\ldots(4)\\ \frac{l}{1}=\frac{m}{2}&=\frac{n}{-1}\ldots (5)\\ \text{From (1) and (4), }&\\ \frac{l}{2}=\frac{m}{-1}&=\frac{n}{-1}\ldots(6)\\ \text{Angle between lines }(5)&\text{ and }(6)\text{is }\\ \cos \theta &=\frac{2-2+1}{\sqrt{6}\sqrt{6}}\\ \cos \theta &=\frac{1}{6}\\ \theta &=\cos^{-1}\left(\frac{1}{6}\right)\end{aligned}\) |
