The angle between the straight lines, whose direction cosines are given by the equations 2l + 2m – n = 0 and mn + nl + lm = 0, is:

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$\frac{π}{3}$

$π - \cos^{- 1} (\frac{4}{9})$

$\cos^{- 1} (\frac{8}{9})$

$\frac{π}{2}$

answer is C.

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Detailed Solution

$\begin{array}{rcl} 2 l + 2 m - n & = & 0 \Rightarrow n = 2 l + 2 m \\ mn + nl + lm & = & 0 \\ m (2 l + 2 m) + (2 l + 2 m) l + lm & = & 0 \\ 2 ml + 2 m^{2} + 2 l^{2} + 2 lm + lm & = & 0 \\ 2 l^{2} + 5 lm + 2 m^{2} & = & 0 \\ (2 l + m) (l + 2 m) & = & 0 \\ m & = & - 2 l, l = - 2 m \\ l_{1} : m_{1} : n_{1} & = & l : - 2 l : - 2 l \\ (l_{1}, m_{1} : n_{1}) & = & (1, - 2, - 2) \\ l_{2} : m_{2} : n_{2} & = & - 2 m : m : - 2 m \\ (l_{2}, m_{2}, n_{2}) & = & (- 2, 1, - 2) \end{array}$

$\begin{array}{rcl} cosθ & = & \frac{l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2}}{\sqrt{{l_{1}}^{2} + {m_{1}}^{2} + {n_{1}}^{2}} \sqrt{{l_{2}}^{2} + {m_{2}}^{2} + {n_{2}}^{2}}} \\ = & \frac{- 2 - 2 + 4}{\sqrt{1 + 4 + 4} \sqrt{4 + 1 + 4}} \\ = & 0 \\ θ & = & \frac{π}{2} \end{array}$