Question

If the square of the shortest distance between the lines $$ \frac{x-2}{1} = \frac{y-1}{2} = \frac{z+3}{-3} \quad \text{and} \quad \frac{x+1}{2} = \frac{y+3}{4} = \frac{z+5}{-5}, $$ is \( \frac{m}{n} \), where \( m, n \) are coprime numbers, then \( m+n \) is equal to: 

• A. 6
• B. 9
• C. 14
• D. 21

Answer 1

The Correct Option is B

The shortest distance between two skew lines is given by: \[ d = \frac{|(\vec{a_1} - \vec{a_2}) \cdot (\vec{b_1} \times \vec{b_2})|}{|\vec{b_1} \times \vec{b_2}|}, \] where \( \vec{a_1}, \vec{a_2} \) are points on each line and \( \vec{b_1}, \vec{b_2} \) are the direction vectors of the lines. Using the given data, calculate the shortest distance and square it to find \( \frac{m}{n} \). The result gives \( m + n = 9 \). Thus, the correct answer is (b) 9.