Question

Two 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 at a point. Then the value of \( k \) is

• A. \( \frac{13}{2} \)
• B. \( -\frac{13}{2} \)
• C. \( \frac{7}{2} \)
• D. \( \frac{9}{2} \)

Hint:

For intersection of two 3D lines, convert both into parametric form and equate corresponding coordinatesSolve for parameters first, then find the unknown constant.

Answer 1

The Correct Option is D

Step 1: Write the first line in parametric form.
Let:
\[ \frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}=t \]
Then:
\[ x=1+2t,\quad y=-1+3t,\quad z=1+4t \]

Step 2: Write the second line in parametric form.
Let:
\[ \frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}=s \]
Then:
\[ x=3+s,\quad y=k+2s,\quad z=s \]

Step 3: Use intersection condition.
Since the lines intersect, their coordinates must be equal at some point.
So:
\[ 1+2t=3+s \]
\[ -1+3t=k+2s \]
\[ 1+4t=s \]

Step 4: Substitute \( s=1+4t \) in x-coordinate equation.
\[ 1+2t=3+(1+4t) \]
\[ 1+2t=4+4t \]
\[ -3=2t \]
\[ t=-\frac{3}{2} \]

Step 5: Find \( s \).
\[ s=1+4t \]
\[ s=1+4\left(-\frac{3}{2}\right) \]
\[ s=1-6=-5 \]

Step 6: Use y-coordinate equation.
\[ -1+3t=k+2s \]
Substitute \( t=-\frac{3}{2} \) and \( s=-5 \):
\[ -1+3\left(-\frac{3}{2}\right)=k+2(-5) \]
\[ -1-\frac{9}{2}=k-10 \]
\[ -\frac{11}{2}=k-10 \]

Step 7: Find \( k \).
\[ k=10-\frac{11}{2} \]
\[ k=\frac{20-11}{2} \]
\[ k=\frac{9}{2} \]
\[ \boxed{\frac{9}{2}} \]