Question

The shortest distance between the lines $\frac{x-5}{1}=\frac{y-2}{2}=\frac{z-4}{-3}$ and $\frac{x+3}{1}=\frac{y+5}{4}=\frac{z-1}{-5}$ is

• A. $7 \sqrt{3}$
• B. $6 \sqrt{3}$
• C. $4 \sqrt{3}$
• D. $5 \sqrt{3}$

Answer 1

The Correct Option is B

The correct answer is option (B) : \(6 \sqrt{3}\)

Shortest distance between two lines
\(\frac{x-x_1}{a_1} = \frac{y-y_1}{a_2} = \frac{z-z_1}{a_3}\) and \(\frac{x-x_2}{b_1} = \frac{y-y_2}{b_2} = \frac{z-z_2}{b_3}\) is given as
\(= \frac{\begin{vmatrix} x_1-x_2 & y_1-y_2 & z_1-z_2\\ a_1 & a_2 & a_3\\ b_1 & b_2 & b_3 \end{vmatrix}}{\sqrt{(a_1b_3-a_3b_2)^2+(a_1b_3-a_3b_1)^2+(a_1b_2-a_2b_1)^2}}\)
\(= \frac{\begin{vmatrix} 5-(3) & 2-(-5) & 4-1\\ 1 & 2 & -3\\ 1 & 4 & -5 \end{vmatrix}}{\sqrt{(-10+12)^2+(-5+3)^2+(4-2)^2}}\)
\(= \frac{\begin{vmatrix} 8 & 7 & 3\\ 1 & 2 & -3\\ 1 & 4 & -5 \end{vmatrix}}{\sqrt{(2)^2+(-2)^2+(2)^2}}\)
\(= \frac{\begin{vmatrix} 8(-10+12)-7(-5+3)+3(4-2)\end{vmatrix}}{\sqrt{4+4+4}}\)
\(= \frac{\begin{vmatrix} 16+14+6\end{vmatrix}}{\sqrt{12}}\)
\(= \frac{36}{\sqrt{12}} = \frac{36}{2\sqrt{3}}\)
\(= \frac{18}{\sqrt{3}} = 6\sqrt{3}\)

Answer 2

Step 1: Shortest distance formula between skew lines: \[ d = \frac{|(a_1 b_2 - a_2 b_1) + (a_2 b_3 - a_3 b_2) + (a_3 b_1 - a_1 b_3)|} {\sqrt{(a_1 - a_2)^2 + (b_1 - b_2)^2 + (c_1 - c_2)^2}} \] 

Step 2: Substituting values, \[ d = \frac{8(-10 + 12) - 7(-5 + 3) + 3(4 - 2)}{\sqrt{4 + 4 + 4}} \] \[ = \frac{16 + 14 + 6}{\sqrt{12}} = \frac{36}{\sqrt{12}} = \frac{36}{2\sqrt{3}} \] \[ = \frac{18}{\sqrt{3}} = 6\sqrt{3} \]