Question

Line $ L_1 $ passes through the point (1, 2, 3) and is parallel to z-axis. Line $ L_2 $ passes through the point $ (\lambda, 5, 6) $ and is parallel to y-axis. Let for $ \lambda = \lambda_1, \lambda_2, \lambda_2<\lambda_1 $, the shortest distance between the two lines be 3. Then the square of the distance of the point $ (\lambda_1, \lambda_2, 7) $ from the line $ L_1 $ is

• A. 40
• B. 32
• C. 25
• D. 37

Hint:

Use the formula for the shortest distance between two skew lines. To find the foot of the perpendicular, use the dot product of the direction vectors.

Answer 1

The Correct Option is C

The equation of line \( L_1 \) is: \( \frac{x-1}{0} = \frac{y-2}{0} = \frac{z-3}{1} \) 

The equation of line \( L_2 \) is: \( \frac{x-\lambda}{0} = \frac{y-5}{1} = \frac{z-6}{0} \) 

The shortest distance (SD) between the lines is given by: \( SD = \frac{ \begin{vmatrix} \lambda - 1 & 5 - 2 & 6 - 3 0 & 0 & 1 \\ 0 & 1 & 0\end{vmatrix} }{ \sqrt{0^2 + 1^2 + 0^2} } \) 

\( SD = \begin{vmatrix} \lambda - 1 & 3 & 3 0 & 0 & 1 \\ 0 & 1 & 0 \end{vmatrix} \) 

\( SD = |\lambda - 1| \) 

Given SD = 3, we have: \( |\lambda - 1| = 3 \) \( \lambda - 1 = \pm 3 \) \( \lambda = 4, -2 \) 

Since \( \lambda_2<\lambda_1 \), we have: \( \lambda_1 = 4 \) \( \lambda_2 = -2 \) The point is \( P(4, -2, 7) \). 

Let Q be the foot of the perpendicular from P to \( L_1 \). Q is of the form \( (1, 2, 3+t) \). 

The direction vector of PQ is \( (1-4, 2-(-2), 3+t-7) = (-3, 4, t-4) \). 

The direction vector of \( L_1 \) is \( (0, 0, 1) \). 

Since PQ is perpendicular to \( L_1 \), their dot product is 0. \( (-3, 4, t-4) \cdot (0, 0, 1) = 0 \) \( t-4 = 0 \) \( t = 4 \) So, Q is \( (1, 2, 7) \). 

Now, \( PQ^2 = (4-1)^2 + (-2-2)^2 + (7-7)^2 \) \( PQ^2 = 3^2 + (-4)^2 + 0^2 \) \( PQ^2 = 9 + 16 = 25 \)