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If the lines $\frac{x-1}{2}=\frac{y+2}{3}=\frac{z-1}{4}$ and…
If the lines $\frac{x-1}{2}=\frac{y+2}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-\mathrm{k}}{2}=\frac{\mathrm{z}}{1}$ intersect, then k has the value
$\frac{7}{2}$ $\frac{3}{2}$ $\frac{-7}{2}$ $\frac{-3}{2}$ Solution
The given equation of lines are $\frac{x-1}{2}=\frac{y+2}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-\mathrm{k}}{2}=\frac{z}{1}$
Since the lines intersect,
$\begin{aligned}
& \quad\left|\begin{array}{ccc}
x_2-x_1 & y_2-y_1 & \mathrm{z}_2-\mathrm{z}_1 \\
\mathrm{a}_1 & \mathrm{~b}_1 & \mathrm{c}_1 \\
\mathrm{a}_2 & \mathrm{~b}_2 & \mathrm{c}_2
\end{array}\right|=0 \\
& \therefore\left|\begin{array}{ccc}
2 & \mathrm{k}+2 & -1 \\
2 & 3 & 4 \\
1 & 2 & 1
\end{array}\right|=0 \\
& \Rightarrow 2(3-8)-(\mathrm{k}+2)(2-4)-1(4-3)=0 \\
& \Rightarrow-10+2 \mathrm{k}+4-1=0 \\
& \Rightarrow 2 \mathrm{k}-7=0 \\
& \Rightarrow \mathrm{k}=\frac{7}{2}
\end{aligned}$
Asked in: MHT CET 2024 (15 May Shift 2)
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