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 given by \( \frac{x-1}{0} = \frac{y-2}{0} = \frac{z-3}{1} \).

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

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

Simplifying this expression yields \( SD = \begin{vmatrix} \lambda - 1 & 3 & 3 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{vmatrix} \).

Further simplification leads to \( SD = |\lambda - 1| \).

Given that the shortest distance \( SD = 3 \), we have the equation \( |\lambda - 1| = 3 \). Solving for \( \lambda \) yields \( \lambda - 1 = \pm 3 \), which gives \( \lambda = 4 \) or \( \lambda = -2 \).

Since it is specified that \( \lambda_2<\lambda_1 \), we assign \( \lambda_1 = 4 \) and \( \lambda_2 = -2 \). The point is \( P(4, -2, 7) \).

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

The direction vector of the line segment PQ is calculated as \( (1-4, 2-(-2), 3+t-7) \), which simplifies to \( (-3, 4, t-4) \).

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

Since PQ is perpendicular to \( L_1 \), their dot product must be zero: \( (-3, 4, t-4) \cdot (0, 0, 1) = 0 \). This implies \( t-4 = 0 \), so \( t = 4 \). Therefore, the coordinates of point Q are \( (1, 2, 7) \).

The square of the distance PQ is calculated as \( PQ^2 = (4-1)^2 + (-2-2)^2 + (7-7)^2 \).

Substituting the values, we get \( PQ^2 = 3^2 + (-4)^2 + 0^2 \).

This results in \( PQ^2 = 9 + 16 = 25 \).