Question

The lines \[ \frac{x - 2}{1} = \frac{y - 3}{1} = \frac{z - 4}{-k} \] and \[ \frac{x - 1}{k} = \frac{y - 4}{2} = \frac{z - 5}{1} \] are coplanar if:

• A. \( k = 3 \) or \( -2 \)
• B. \( k = 0 \) or \( -1 \)
• C. \( k = 1 \) or \( -1 \)
• D. \( k = 0 \) or \( -3 \)

Hint:

For coplanar lines, use the scalar triple product of the direction ratios and the vector joining points on the lines. If the product is zero, the lines are coplanar.

Answer 1

The Correct Option is D

Step 1: To check if two lines are coplanar, we use the condition that the scalar triple product of the direction vectors of the lines and the vector joining a point on one line to a point on the other line must be zero. 
The direction ratios of the first line are: \( (1, 1, -k) \), and for the second line, the direction ratios are: \( (k, 2, 1) \). 
The point on the first line can be taken as \( (2, 3, 4) \), and the point on the second line can be taken as \( (1, 4, 5) \). 
The vector joining these two points is: \[ (1 - 2, 4 - 3, 5 - 4) = (-1, 1, 1). \] 
We now calculate the scalar triple product: \[ \left| \begin{array}{ccc} 1 & 1 & -k \\ k & 2 & 1 \\ -1 & 1 & 1 \end{array} \right| = 1 \cdot \left| \begin{array}{cc} 2 & 1 \\ 1 & 1 \end{array} \right| - 1 \cdot \left| \begin{array}{cc} k & 1 \\ -1 & 1 \end{array} \right| + (-k) \cdot \left| \begin{array}{cc} k & 2 \\ -1 & 1 \end{array} \right| \] \[ = 1 \cdot (2 \cdot 1 - 1 \cdot 1) - 1 \cdot (k \cdot 1 - (-1) \cdot 1) + (-k) \cdot (k \cdot 1 - (-1) \cdot 2) \] \[ = 1 \cdot 1 - 1 \cdot (k + 1) - k \cdot (k + 2) \] \[ = 1 - (k + 1) - k(k + 2) \] \[ = 1 - k - 1 - k^2 - 2k = -k^2 - 3k. \] 
For the lines to be coplanar, the scalar triple product must be zero: \[ -k^2 - 3k = 0 \] \[ k(k + 3) = 0. \] Thus, \( k = 0 \) or \( k = -3 \).