MHT CET · Maths · Three Dimensional Geometry
If the lines \(\frac{x+1}{-10}=\frac{y+\mathrm{k}}{-1}=\frac{z-4}{1}\) and \(\frac{x+10}{-1}=\frac{y+1}{-3}=\frac{z-1}{4}\) intersect each other, then the value of \(k\) is
- A -3
- B 3
- C 4
- D 2
Answer & Solution
Correct Answer
(B) 3
Step-by-step Solution
Detailed explanation
Since the given lines intersect,
\(\begin{aligned}
& \left|\begin{array}{ccc}
x_2-x_1 & y_2-y_1 & z_2-z_1 \\
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2
\end{array}\right|=0 \\
& \left|\begin{array}{ccc}
-10+1 & -1+\mathrm{k} & 1-4 \\
-10 & -1 & 1 \\
-1 & -3 & 4
\end{array}\right|=0 \\
& \Rightarrow\left|\begin{array}{ccc}
-9 & -1+\mathrm{k} & -3 \\
-10 & -1 & 1 \\
-1 & -3 & 4
\end{array}\right|=0 \\
& \Rightarrow 39 \mathrm{k}=117 \\
& \Rightarrow \mathrm{k}=3
\end{aligned}\)
\(\begin{aligned}
& \left|\begin{array}{ccc}
x_2-x_1 & y_2-y_1 & z_2-z_1 \\
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2
\end{array}\right|=0 \\
& \left|\begin{array}{ccc}
-10+1 & -1+\mathrm{k} & 1-4 \\
-10 & -1 & 1 \\
-1 & -3 & 4
\end{array}\right|=0 \\
& \Rightarrow\left|\begin{array}{ccc}
-9 & -1+\mathrm{k} & -3 \\
-10 & -1 & 1 \\
-1 & -3 & 4
\end{array}\right|=0 \\
& \Rightarrow 39 \mathrm{k}=117 \\
& \Rightarrow \mathrm{k}=3
\end{aligned}\)
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