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 the value of $k$ is

• A. $\frac{3}{2}$
• B. $-\frac{3}{2}$
• C. $-\frac{2}{9}$
• D. $\frac{9}{2}$

Hint:

Alternatively, find any general point on Line 1: $(2\lambda + 1, 3\lambda - 1, 4\lambda + 1)$. Equating the z-coordinate to Line 2's parameter ($\mu$) via $4\lambda + 1 = \mu$ and x-coordinate $2\lambda + 1 = \mu + 3$ lets you solve for $\lambda = -\frac{3}{2}$ and $\mu = -5$ in seconds. Plugging these into the y-coordinate equations provides a seamless double check for $k = \frac{9}{2}$!

Answer 1

The Correct Option is D

Step 1: Set up coplanarity.
Line 1 has point $(1,-1,1)$ and direction $(2,3,4)$. Line 2 has point $(3,k,0)$ and direction $(1,2,1)$. For intersection the determinant of the joining vector and the two directions is zero.

Step 2: Write the determinant.
The joining vector is $(2, k+1, -1)$, so \[ \begin{vmatrix} 2 & k+1 & -1 \\ 2 & 3 & 4 \\ 1 & 2 & 1 \end{vmatrix} = 0 \]

Step 3: Expand and solve.
This gives $-10 + 2(k+1) - 1 = 0$, so $2k - 9 = 0$ and $k = \dfrac{9}{2}$.
\[ \boxed{\dfrac{9}{2},\ \text{option 4}} \]