The shortest distance between the line x - 3 / 4 = y + 7 / -11 = z - 1 / 5 and x - 5 / 3 = y - 9 / -6

Question

The shortest distance between the line \(\frac{x-3}{4}=\frac{y+7}{-11}=\frac{z-1}{5}\) and \(\frac{x-5}{3}=\frac{y-9}{-6}=\frac{z+2}{1}\) is :

(1) \(\frac{187}{\sqrt{563}}\)

(2) \(\frac{178}{\sqrt{563}}\)

(3) \(\frac{185}{\sqrt{563}}\)

(4) \(\frac{179}{\sqrt{563}}\) 

Answer 1

Correct option is : (1) \(\frac{187}{\sqrt{563}}\) 

\(\vec{n}=\vec{p} \times \vec{q}\)

\(\vec{n}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 4 & -11 & 5 \\ 3 & -6 & 1\end{array}\right|=19 \hat{i}+11 \hat{j}+9 \hat{k}\)

S.d. = projection of \(\overrightarrow{\mathrm{AB}}\) on \(\vec{n}\)

\(=\left|\frac{\overrightarrow{\mathrm{AB}} \cdot \overrightarrow{\mathrm{n}}}{|\overrightarrow{\mathrm{n}}|}\right|=\left|\frac{(2 \hat{i}+16 \hat{j}-3 \hat{k}) \cdot(19 \hat{i}+11 \hat{j}+9 \hat{k})}{\sqrt{361+121+81}}\right|\)

\(=\frac{38+176-27}{\sqrt{563}}\)

S.d. \(=\frac{187}{\sqrt{563}}\)