Question

If two lines $ L_1 : \frac{x-1}{2} = \frac{y+1}{3} = \frac{z-1}{4} $ and $ L_2 : \frac{x-3}{1} = \frac{y-k}{2} = z $ intersect at a point, then $ 2k $ is equal to

• A. 9
• B. \( \frac{1}{2} \)
• C. \( \frac{9}{2} \)
• D. 1

Hint:

For two lines to intersect, solve the system of parametric equations and check for the consistency of the results.

Answer 1

The Correct Option is A

The condition for two lines to intersect is that their parametric equations must give the same value at the point of intersection. For the line \( L_1 \), we can write the parametric equations: \[ x_1 = 1 + 2t, \quad y_1 = -1 + 3t, \quad z_1 = 1 + 4t \] For the line \( L_2 \), we can write the parametric equations: \[ x_2 = 3 + s, \quad y_2 = k + 2s, \quad z_2 = s \] At the point of intersection, the coordinates must be equal, so: \[ x_1 = x_2, \quad y_1 = y_2, \quad z_1 = z_2 \] This gives us the following system of equations: 1. \( 1 + 2t = 3 + s \) 2. \( -1 + 3t = k + 2s \) 3. \( 1 + 4t = s \) From equation (3), we can solve for \( s \): \[ s = 1 + 4t \] Substitute this into equation (1): \[ 1 + 2t = 3 + (1 + 4t) \] Simplify: \[ 1 + 2t = 4 + 4t \] \[ -3 = 2t \] \[ t = -\frac{3}{2} \] Now substitute \( t = -\frac{3}{2} \) into the equation for \( s \): \[ s = 1 + 4\left(-\frac{3}{2}\right) = 1 - 6 = -5 \] Substitute \( t = -\frac{3}{2} \) and \( s = -5 \) into equation (2) to find \( k \): \[ -1 + 3\left(-\frac{3}{2}\right) = k + 2(-5) \] \[ -1 - \frac{9}{2} = k - 10 \] \[ -\frac{11}{2} = k - 10 \] \[ k = \frac{9}{2} \] Thus, \( 2k = 9 \).