JEE MAIN - Mathematics (2024 - 8th April Evening Shift - No. 2)

If the shortest distance between the lines $$\frac{x-\lambda}{2}=\frac{y-4}{3}=\frac{z-3}{4}$$ and $$\frac{x-2}{4}=\frac{y-4}{6}=\frac{z-7}{8}$$ is $$\frac{13}{\sqrt{29}}$$, then a value of $$\lambda$$ is :
$$\frac{13}{25}$$
1
$$-$$1
$$-\frac{13}{25}$$

Explanation

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

JEE Main 2024 (Online) 8th April Evening Shift Mathematics - 3D Geometry Question 29 English Explanation

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

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

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

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