Q47 / 75 · JEE Main 2025 (2 Apr Shift 2) · Mathematics
The line $L_1$ is parallel to the vector $\vec{a}=-3 \hat{i}+2 \hat{j}+4 \hat{k}$ and passes through the point $(7,6,2)$ and the line $L_2$ is parallel to the vector $\vec{b}=2 \hat{i}+\hat{j}+3 \hat{k}$ and passes through the point $(5,3,4)$. The shortest distance between the lines $L_1$ and $L_2$ is :
- Correct answerA$\frac{23}{\sqrt{38}}$
- B$\frac{21}{\sqrt{57}}$
- C$\frac{23}{\sqrt{57}}$
- D$\frac{21}{\sqrt{38}}$
Solution
$\begin{aligned} & L_1:(7 \hat{i}+6 \hat{j}+2 k)+\lambda(-3 \hat{i}+2 \hat{j}+4 k) \\ & L_2:(5 \hat{i}+3 \hat{j}+4 k)+\lambda(2 \hat{i}+\hat{j}+3 k)\end{aligned}$
Distance between skew lines
$\begin{aligned} & =\frac{(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}) \cdot(2 \hat{\mathrm{i}}+17 \hat{\mathrm{j}}-7 \hat{\mathrm{k}})}{\sqrt{342}} \\ & =\frac{69}{\sqrt{342}}=\frac{69}{3 \sqrt{38}}=\frac{23}{\sqrt{38}}\end{aligned}$
Distance between skew lines
$\begin{aligned} & =\frac{(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}) \cdot(2 \hat{\mathrm{i}}+17 \hat{\mathrm{j}}-7 \hat{\mathrm{k}})}{\sqrt{342}} \\ & =\frac{69}{\sqrt{342}}=\frac{69}{3 \sqrt{38}}=\frac{23}{\sqrt{38}}\end{aligned}$
Hence, the correct answer is option: 'A'.