Question

If the lines \( \frac{x-1}{2} = \frac{y+1}{3} = \frac{z-1}{4} \) and \( \frac{x-3}{1} = \frac{y-k}{2} = \frac{z}{1} \) intersect, then \(k\) is equal to:

• A. \(-1\)
• B. \( \frac{2}{9} \)
• C. \( \frac{9}{2} \)
• D. \(0\)

Hint:

For intersecting lines → equate coordinates and solve parameters systematically.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
If two lines in 3D space intersect, there exists a unique point that satisfies the equations of both lines simultaneously. We find generic points on both lines and equate their coordinates.
Step 3: Detailed Explanation:
1. Points on Line 1: Let \(\frac{x-1}{2} = \frac{y+1}{3} = \frac{z-1}{4} = p\).
Generically, \(P = (2p+1, 3p-1, 4p+1)\).
2. Points on Line 2: Let \(\frac{x-3}{1} = \frac{y-k}{2} = \frac{z}{1} = q\).
Generically, \(Q = (q+3, 2q+k, q)\).
3. For intersection, \(P = Q\). Equate x and z:
\[ 2p + 1 = q + 3 \implies 2p - q = 2 \dots (i) \]
\[ 4p + 1 = q \dots (ii) \]
4. Substitute \(q\) from (ii) into (i):
\[ 2p - (4p + 1) = 2 \implies -2p - 1 = 2 \implies -2p = 3 \implies p = -1.5 \]
5. Find \(q\): \(q = 4(-1.5) + 1 = -6 + 1 = -5\).
6. Equate y coordinates using found \(p\) and \(q\):
\[ 3p - 1 = 2q + k \]
\[ 3(-1.5) - 1 = 2(-5) + k \]
\[ -4.5 - 1 = -10 + k \implies -5.5 = -10 + k \implies k = 4.5 \]
Converting to fraction: \(4.5 = 9/2\).
Step 4: Final Answer:
The value of \(k\) is 9/2.