If the D.r's of two lines are given by l + m + n = 0, mn - 2ln + lm = 0 then the anlge betweenthe lines is

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$\frac{π}{3}$

$\frac{π}{4}$

$\frac{π}{2}$

answer is C.

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Detailed Solution

$\begin{array}{rcl} l + m + n & = & 0 \Rightarrow l = - m - n \\ mn - 2 \ln + lm & = & 0 \\ mn + 2 (m + n) n - (m + n) m & = & 0 \\ mn + 2 mn + 2 n^{2} - m^{2} - mn & = & 0 \\ 2 n^{2} + 2 mn - m^{2} & = & 0 \\ {(\frac{m}{n})}^{2} - 2 \frac{m}{n} - 2 & = & 0 \end{array}$

$\frac{m_{1}}{n_{1}}, \frac{m_{2}}{n_{2}}$ are roots

$\begin{array}{rcl} \frac{m_{1}}{n_{1}}, \frac{m_{2}}{n_{2}} & = & - 2 \\ m & = & - l - n \\ mn - 2 \ln + lm & = & 0 \\ - (l + n) n - 2 \ln + l (- l - n) & = & 0 \\ - \ln - n^{2} - 2 \ln - l^{2} - \ln & = & 0 \\ l^{2} + 4 \ln + n^{2} & = & 0 \\ {(\frac{l}{n})}^{2} + 4 \frac{l}{n} + 1 & = & 0 \end{array}$

$\frac{l_{1}}{n_{1}}, \frac{l_{2}}{n_{2}}$ are roots

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