If lines frac{2x-4}{lambda}=frac{y-1}{2}=frac | vedclass

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(D) Key Idea: If lines $\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}$ and $\frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}$ are perpend

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$-\frac{6}{7}$

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(D) Key Idea: If lines $\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}$ and $\frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}$ are perpendicular,then $a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$.Given lines are:$\frac{2x-4}{\lambda} = \frac{y-1}{2} = \frac{z-3}{1} \Rightarrow \frac{x-2}{\lambda/2} = \frac{y-1}{2} = \frac{z-3}{1}$ $(i)$And $\frac{x-1}{1} = \frac{3y-1}{\lambda} = \frac{z-2}{1} \Rightarrow \frac{x-1}{1} = \frac{y-1/3}{\lambda/3} = \frac{z-2}{1}$ (ii)Since lines $(i)$ and (ii) are perpendicular,the dot product of their direction ratios must be zero:$(\frac{\lambda}{2})(1) + (2)(\frac{\lambda}{3}) + (1)(1) = 0$$\frac{\lambda}{2} + \frac{2\lambda}{3} + 1 = 0$Multiplying by $6$ to clear the denominators:$3\lambda + 4\lambda + 6 = 0$$7\lambda = -6$$\lambda = -\frac{6}{7}$

If lines $\frac{2x-4}{\lambda}=\frac{y-1}{2}=\frac{z-3}{1}$ and $\frac{x-1}{1}=\frac{3y-1}{\lambda}=\frac{z-2}{1}$ are perpendicular to each other,then $\lambda = \ldots$.

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