MHT CET · Maths · Three Dimensional Geometry
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
- A \(\frac {1}{2}\)
- B \(-1\)
- C \(1\)
- D \(\frac {3}{2}\)
Answer & Solution
Correct Answer
(C) \(1\)
Step-by-step Solution
Detailed explanation
Since the given lines intersect
\(\begin{aligned}
&\left|\begin{array}{ccc}
x_2-x_1 & y_2-y_1 & \mathrm{z}_2-\mathrm{z}_1 \\
\mathrm{a}_1 & \mathrm{~b}_1 & \mathrm{c}_1 \\
\mathrm{a}_2 & \mathrm{~b}_2 & \mathrm{c}_2
\end{array}\right|=0 \\
& \therefore \quad\left|\begin{array}{ccc}
3-\mathrm{k} & \frac{9}{2}+1 & 0-1 \\
2 & 3 & 4 \\
1 & 2 & 1
\end{array}\right|=0 \\
& \Rightarrow\left|\begin{array}{ccc}
3-\mathrm{k} & \frac{11}{2} & -1 \\
2 & 3 & 4 \\
1 & 2 & 1
\end{array}\right|=0 \\
& \Rightarrow-5+5 \mathrm{k}=0 \\
& \Rightarrow \mathrm{k}=1
\end{aligned}\)
\(\begin{aligned}
&\left|\begin{array}{ccc}
x_2-x_1 & y_2-y_1 & \mathrm{z}_2-\mathrm{z}_1 \\
\mathrm{a}_1 & \mathrm{~b}_1 & \mathrm{c}_1 \\
\mathrm{a}_2 & \mathrm{~b}_2 & \mathrm{c}_2
\end{array}\right|=0 \\
& \therefore \quad\left|\begin{array}{ccc}
3-\mathrm{k} & \frac{9}{2}+1 & 0-1 \\
2 & 3 & 4 \\
1 & 2 & 1
\end{array}\right|=0 \\
& \Rightarrow\left|\begin{array}{ccc}
3-\mathrm{k} & \frac{11}{2} & -1 \\
2 & 3 & 4 \\
1 & 2 & 1
\end{array}\right|=0 \\
& \Rightarrow-5+5 \mathrm{k}=0 \\
& \Rightarrow \mathrm{k}=1
\end{aligned}\)
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