The shortest distance between lines L1 and L2, where L1 : x−1/2 = y+1/−3 = z+4/2

Question

The shortest distance between lines \(\mathrm{L}_{1}\) and \(\mathrm{L}_{2}\), where \(\mathrm{L}_{1}: \frac{\mathrm{x}-1}{2}=\frac{\mathrm{y}+1}{-3}=\frac{\mathrm{z}+4}{2}\) and \(\mathrm{L}_{2}\) is the line passing through the points \(A(-4,4,3) \cdot B(-1,6,3)\) and perpendicular to the line \(\frac{x-3}{-2}=\frac{y}{3}=\frac{z-1}{1}\), is

(1) \(\frac{121}{\sqrt{221}}\)

(2) \(\frac{24}{\sqrt{117}}\)

(3) \(\frac{141}{\sqrt{221}}\)

(4) \(\frac{42}{\sqrt{117}}\)

Answer 1

Correct option is (3) \(\frac{141}{\sqrt{221}}\)

\(\mathrm{L}_{2}=\frac{\mathrm{x}+4}{3}=\frac{\mathrm{y}-4}{2}=\frac{\mathrm{z}-3}{0}\)

\( \therefore \text { S.D }=\frac{\left|\begin{array}{ccc} \mathrm{x}_{2}-\mathrm{x}_{1} & \mathrm{y}_{2}-\mathrm{y}_{1} & \mathrm{z}_{2}-\mathrm{z}_{1} \\ 2 & -3 & 2 \\ 3 & 2 & 0 \end{array}\right|}{\left|\overrightarrow{\mathrm{n}_{1}} \times \overrightarrow{\mathrm{n}_{2}}\right|} \)

\(=\frac{\left|\begin{array}{ccc} 5 & -5 & -7 \\ 2 & -3 & 2 \\ 3 & 2 & 0 \end{array}\right|}{\left|\overrightarrow{\mathrm{n}_{1}} \times \overrightarrow{\mathrm{n}_{2}}\right|} \)

\(=\frac{141}{|-4 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}+13 \hat{\mathrm{k}}|}\)

\(=\frac{141}{\sqrt{16+36+169}}\)

\(=\frac{141}{\sqrt{221}}\)