If the lines $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4} \quad$ and $x-3=\frac{y-\mathrm{k}}{2}=\mathrm{z}$…

If the lines $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4} \quad$ and $x-3=\frac{y-\mathrm{k}}{2}=\mathrm{z}$ intersect, then the value of $\mathrm{k}$ is
  1. $\frac{3}{2}$
  2. $\frac{-2}{9}$
  3. $\frac{-2}{3}$
  4. $\frac{9}{2}$

Solution

As the given lines are intersecting, the shortest distance between them is zero. $\begin{array}{ll} \therefore & \left|\begin{array}{ccc} x_2-x_1 & y_2-y_1 & \mathrm{z}_2-\mathrm{z}_1 \\ \mathrm{a}_1 & \mathrm{~b}_1 & \mathrm{c}_1 \\ \mathrm{a}_2 & \mathrm{~b}_2 & \mathrm{c}_2 \end{array}\right|=0 \\ \therefore & \left|\begin{array}{ccc} 3-1 & \mathrm{k}+1 & 0-1 \\ 2 & 3 & 4 \\ 1 & 2 & 1 \end{array}\right|=0 \\ \therefore & \left|\begin{array}{ccc} 2 & \mathrm{k}+1 & -1 \\ 2 & 3 & 4 \\ 1 & 2 & 1 \end{array}\right|=0 \\ \therefore & \mathrm{k}=\frac{9}{2} \end{array}$

Asked in: MHT CET 2023 (11 May Shift 2)

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