QJEEM24Ma5501: JEE Main 2024 Mathematics - Class 12 JEEM24Ma5501 Solution: If the shortest distance between the lines and is , then a…

$\begin{aligned}& \vec{l}_1=2 \hat{i}+3 \hat{j}+4 \hat{k} \\& \vec{l}_2=4 \hat{i}+6 \hat{j}+8 \hat{k}\end{aligned}$

$S . D .=\frac{|(2 \hat{i}+3 \hat{j}+4 \hat{k}) \times((\lambda-2) \hat{i}-4 \hat{k})|}{|2 \hat{i}+3 \hat{j}+4 \hat{k}|}$

$\begin{aligned}& \left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\2 & 3 & 4 \\\lambda-2 & 0 & -4\end{array}\right|=-12 \hat{i}-\hat{j}(-8-4 \lambda+8)+\hat{k}(6-3 \lambda) \\& =-12 \hat{i}+4 \lambda \hat{j}+(6-3 \lambda) \hat{k} \\& \frac{\sqrt{144+16 \lambda^2+(6-3 \lambda)^2}}{\sqrt{29}}=\frac{13}{\sqrt{29}} \\& 144+16 \lambda^2+(6-3 \lambda)^2=169 \\& \Rightarrow 16 \lambda^2+9 \lambda^2+36-36 \lambda+144-169=0 \\\end{aligned}$

$\begin{aligned}& \Rightarrow 25 \lambda^2-36 \lambda+11=0 \\& \Rightarrow 25 \lambda^2-25 \lambda-11 \lambda+11=0 \\& \Rightarrow(25 \lambda-11)(\lambda-1)=0 \\& \Rightarrow \lambda=1, \frac{11}{25}\end{aligned}$