JEE MAIN - Mathematics (2024 - 31st January Evening Shift - No. 8)
The shortest distance, between lines $$L_1$$ and $$L_2$$, where $$L_1: \frac{x-1}{2}=\frac{y+1}{-3}=\frac{z+4}{2}$$ and $$L_2$$ is the line, passing through the points $$\mathrm{A}(-4,4,3), \mathrm{B}(-1,6,3)$$ and perpendicular to the line $$\frac{x-3}{-2}=\frac{y}{3}=\frac{z-1}{1}$$, is
$$\frac{141}{\sqrt{221}}$$
$$\frac{24}{\sqrt{117}}$$
$$\frac{42}{\sqrt{117}}$$
$$\frac{121}{\sqrt{221}}$$
Explanation
$$\begin{aligned}
& \mathrm{L}_2=\frac{\mathrm{x}+4}{3}=\frac{\mathrm{y}-4}{2}=\frac{\mathrm{z}-3}{0} \\
& \therefore \mathrm{S} . \mathrm{D}=\frac{\left|\begin{array}{ccc}
\mathrm{x}_2-\mathrm{x}_1 & \mathrm{y}_2-\mathrm{y}_1 & \mathrm{z}_2-\mathrm{z}_1 \\
2 & -3 & 2 \\
3 & 2 & 0
\end{array}\right|}{\left|\overrightarrow{\mathrm{n}_1} \times \overrightarrow{\mathrm{n}_2}\right|} \\
& =\frac{\left|\begin{array}{ccc}
5 & -5 & -7 \\
2 & -3 & 2 \\
3 & 2 & 0
\end{array}\right|}{\left|\overrightarrow{\mathrm{n}_1} \times \overrightarrow{\mathrm{n}_2}\right|} \\
& =\frac{141}{|-4 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}+13 \hat{\mathrm{k}}|} \\
& =\frac{141}{\sqrt{16+36+169}} \\
& =\frac{141}{\sqrt{221}}
\end{aligned}$$
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