Question

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

• A. \(\dfrac{3}{2}\)
• B. \(\dfrac{9}{2}\)
• C. \(\dfrac{2}{9}\)
• D. \(\dfrac{3}{2}\)

Hint:

To check intersection of two lines in 3D, write parametric form and equate \(x,y,z\). Solve parameters and use remaining equation to find unknown constant.

Answer 1

The Correct Option is B

Step 1: Write parametric form of first line.
Let parameter \(t\):
\[ x=1+2t,\quad y=-1+3t,\quad z=1+4t \] Step 2: Write parametric form of second line.
Let parameter \(s\):
\[ x=3+s,\quad y=k+2s,\quad z=s \] Step 3: At intersection, coordinates are equal.
From \(z\):
\[ 1+4t=s \] From \(x\):
\[ 1+2t=3+s \Rightarrow 1+2t=3+(1+4t) \Rightarrow 1+2t=4+4t \Rightarrow -3=2t \Rightarrow t=-\frac{3}{2} \] Then:
\[ s=1+4t=1+4\left(-\frac{3}{2}\right)=1-6=-5 \] Step 4: Use \(y\) equality to find \(k\).
First line:
\[ y=-1+3t=-1+3\left(-\frac{3}{2}\right)=-1-\frac{9}{2}=-\frac{11}{2} \] Second line:
\[ y=k+2s=k+2(-5)=k-10 \] Equate:
\[ k-10=-\frac{11}{2} \Rightarrow k=10-\frac{11}{2}=\frac{20-11}{2}=\frac{9}{2} \] Final Answer: \[ \boxed{\frac{9}{2}} \]