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

To determine the value of \(2k\), we start by finding the point of intersection of the two lines \(L_1\) and \(L_2\).

Step 1: Parametrize \(L_1\)
The equations for line \(L_1\) are given as: \[\frac{x-1}{2} = \frac{y+1}{3} = \frac{z-1}{4} = t\] which provides the parametric equations: \[x = 2t + 1, \quad y = 3t - 1, \quad z = 4t + 1\]

Step 2: Parametrize \(L_2\)
For line \(L_2\): \[\frac{x-3}{1} = \frac{y-k}{2} = z = s\] leading to the parametric equations: \[x = s + 3, \quad y = 2s + k, \quad z = s\]

Step 3: Set equations for intersection
For the lines to intersect, there must exist parameters \(t\) and \(s\) such that:

• \(2t + 1 = s + 3\)
• \(3t - 1 = 2s + k\)
• \(4t + 1 = s\)

From the third equation: \(s = 4t + 1\).

Substitute \(s\) into the first equation:
\[2t + 1 = 4t + 1 + 3\] which simplifies to: \[2t + 1 = 4t + 4\]
Simplifying further gives: \[-2t = 3 \quad \therefore t = -\frac{3}{2}\]

Substitute \(t\) back into the expression for \(s\):
\[s = 4(-\frac{3}{2}) + 1 = -6 + 1 = -5\]

In the second line equation:
\(3t - 1 = 2s + k\), substitute \(t = -\frac{3}{2}\) and \(s = -5\):

\[3(-\frac{3}{2}) - 1 = 2(-5) + k\]
Simplifying gives: \[-\frac{9}{2} - 1 = -10 + k\]

Further simplification: \[-\frac{11}{2} = -10 + k\]

Solve for \(k\):
\[k = -\frac{11}{2} + 10 = -\frac{11}{2} + \frac{20}{2} = \frac{9}{2}\]

Step 4: Calculate \(2k\)
Finally, multiply \(k\) by 2:
\[2k = 2 \times \frac{9}{2} = 9\]

Thus, \(2k\) is equal to 9.

Answer 2

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, \( 2k = 9 \).