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

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

Solution

Since the given lines intersect $\begin{aligned} &\left|\begin{array}{ccc} x_2-x_1 & y_2-y_1 & \mathrm{z}_2-\mathrm{z}_1 \\ \mathrm{a}_1 & \mathrm{~b}_1 & \mathrm{c}_1 \\ \mathrm{a}_2 & \mathrm{~b}_2 & \mathrm{c}_2 \end{array}\right|=0 \\ & \therefore \quad\left|\begin{array}{ccc} 3-\mathrm{k} & \frac{9}{2}+1 & 0-1 \\ 2 & 3 & 4 \\ 1 & 2 & 1 \end{array}\right|=0 \\ & \Rightarrow\left|\begin{array}{ccc} 3-\mathrm{k} & \frac{11}{2} & -1 \\ 2 & 3 & 4 \\ 1 & 2 & 1 \end{array}\right|=0 \\ & \Rightarrow-5+5 \mathrm{k}=0 \\ & \Rightarrow \mathrm{k}=1 \end{aligned}$

Asked in: MHT CET 2023 (14 May Shift 2)

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