Question

If the lines $\frac{x-k}{2} = \frac{y+1}{3} = \frac{z-1}{4}$ and $\frac{x-3}{1} = \frac{y-\frac{9}{2{2} = \frac{z}{1}$ intersect, then the value of k is

• A. $\frac{1}{2}$
• B. $-1$
• C. $1$
• D. $\frac{3}{2}$

Hint:

For two lines $\frac{x-x_1}{a_1} = \frac{y-y_1}{b_1} = \frac{z-z_1}{c_1}$ and $\frac{x-x_2}{a_2} = \frac{y-y_2}{b_2} = \frac{z-z_2}{c_2}$ to intersect, the condition is $\begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix} = 0$.

Answer 1

The Correct Option is A

Step 1: Understand the condition for intersection of two lines.
Two lines intersect if the shortest distance between them is zero. This can be expressed using the scalar triple product. Let the first line $L_1$ pass through $P_1(k, -1, 1)$ with direction vector: \[ \vec{d_1} = (2, 3, 4) \] Let the second line $L_2$ pass through $P_2\left(3, \frac{9}{2}, 0\right)$ with direction vector: \[ \vec{d_2} = (1, 2, 1) \] \[ \vec{P_1P_2} = P_2 - P_1 = \left(3-k, \frac{11}{2}, -1\right) \] For intersection: \[ \begin{vmatrix} 3-k & \frac{11}{2} & -1 \\ 2 & 3 & 4 \\ 1 & 2 & 1 \end{vmatrix} = 0 \]
Step 2: Expand the determinant.
\[ (3-k) \begin{vmatrix} 3 & 4 \\ 2 & 1 \end{vmatrix} - \frac{11}{2} \begin{vmatrix} 2 & 4 \\ 1 & 1 \end{vmatrix} + (-1) \begin{vmatrix} 2 & 3 \\ 1 & 2 \end{vmatrix} = 0 \] \[ (3-k)(3 \cdot 1 - 4 \cdot 2) - \frac{11}{2}(2 \cdot 1 - 4 \cdot 1) - (2 \cdot 2 - 3 \cdot 1) = 0 \] \[ (3-k)(-5) - \frac{11}{2}(-2) - (4 - 3) = 0 \] \[ -15 + 5k + 11 - 1 = 0 \] 
Step 3: Solve for $k$.
\[ 5k - 5 = 0 \] \[ k = 1 \] 
Final Answer:
\[ \boxed{k = 1} \]