MHT CET · Maths · Three Dimensional Geometry
If the line \(\frac{x-1}{-3}=\frac{y-2}{2 k}=\frac{z-3}{2}\) and \(\frac{x-1}{3 k}=\frac{y-5}{1}=\frac{z-6}{-5}\) are perpendicular to each other, then \(k\) is
- A \(\frac{7}{10}\)
- B \(\frac{10}{7}\)
- C \(\frac{-7}{10}\)
- D \(\frac{-10}{7}\)
Answer & Solution
Correct Answer
(D) \(\frac{-10}{7}\)
Step-by-step Solution
Detailed explanation
Given lines are perpendicular. Hence we write
\(
\begin{aligned}
&(-3)(3 \mathrm{k})+(2 \mathrm{k})(1)+(2)(-5)=0 \\
\therefore &-9 \mathrm{k}+2 \mathrm{k}-10=0 \\
\therefore & 7 \mathrm{k}=-10 \Rightarrow \quad \mathrm{k}=\frac{-10}{7}
\end{aligned}
\)
\(
\begin{aligned}
&(-3)(3 \mathrm{k})+(2 \mathrm{k})(1)+(2)(-5)=0 \\
\therefore &-9 \mathrm{k}+2 \mathrm{k}-10=0 \\
\therefore & 7 \mathrm{k}=-10 \Rightarrow \quad \mathrm{k}=\frac{-10}{7}
\end{aligned}
\)
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