MHT CET · Maths · Three Dimensional Geometry
If the line \(\frac{1-x}{3}=\frac{7 y-14}{2 \lambda}=\frac{z-3}{2}\) and \(\frac{7-7 x}{3 \lambda}=\frac{y-5}{1}=\frac{6-z}{5}\) are at right angles, then \(\lambda=\)
- A \(\frac{-70}{11}\)
- B \(\frac{70}{11}\)
- C \(\frac{11}{70}\)
- D \(\frac{-11}{70}\)
Answer & Solution
Correct Answer
(B) \(\frac{70}{11}\)
Step-by-step Solution
Detailed explanation
We have lines
\(
\frac{x-1}{-3}=\frac{7(y-2)}{2 \lambda}=\frac{z-3}{2} \text { and } \frac{7(1-x)}{3 \lambda}=\frac{y-5}{1}=\frac{z-6}{-5}
\)
i.e. \(\frac{x-1}{-3}=\frac{y-2}{\left(\frac{2 \lambda}{7}\right)}=\frac{z-3}{2}\) and \(\frac{x-1}{\left(\frac{-3 \lambda}{7}\right)}=\frac{y-5}{1}=\frac{z-6}{5}\)
Since given lines are at right angles, we write
\(
\begin{aligned}
& (-3)\left(\frac{-3 \lambda}{7}\right)+\left(\frac{2 \lambda}{7}\right)(1)+(2)(-5)=0 \\
& \therefore \frac{9 \lambda}{7}+\frac{2 \lambda}{7}-10=0 \Rightarrow 11 \lambda=70 \Rightarrow \lambda=\frac{70}{11}
\end{aligned}
\)
\(
\frac{x-1}{-3}=\frac{7(y-2)}{2 \lambda}=\frac{z-3}{2} \text { and } \frac{7(1-x)}{3 \lambda}=\frac{y-5}{1}=\frac{z-6}{-5}
\)
i.e. \(\frac{x-1}{-3}=\frac{y-2}{\left(\frac{2 \lambda}{7}\right)}=\frac{z-3}{2}\) and \(\frac{x-1}{\left(\frac{-3 \lambda}{7}\right)}=\frac{y-5}{1}=\frac{z-6}{5}\)
Since given lines are at right angles, we write
\(
\begin{aligned}
& (-3)\left(\frac{-3 \lambda}{7}\right)+\left(\frac{2 \lambda}{7}\right)(1)+(2)(-5)=0 \\
& \therefore \frac{9 \lambda}{7}+\frac{2 \lambda}{7}-10=0 \Rightarrow 11 \lambda=70 \Rightarrow \lambda=\frac{70}{11}
\end{aligned}
\)
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