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: Understanding the Question:
We are given two lines in three-dimensional space expressed in symmetrical Cartesian form. We are told that these lines intersect at a common point, and we need to determine the value of the unknown constant $k$.

Step 2: Key Formula or Approach:
For two lines to intersect, they must be coplanar. If a line passes through point $A(x_1, y_1, z_1)$ with direction ratios $(a_1, b_1, c_1)$ and another line passes through point $B(x_2, y_2, z_2)$ with direction ratios $(a_2, b_2, c_2)$, the condition for intersection is given by setting the coplanarity determinant to zero: $$ \begin{vmatrix} x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix} = 0 $$

Step 3: Detailed Explanation:
Let's extract the coordinates and direction components from the two lines:

• Line 1 passes through $A(1, -1, 1)$ with direction vector $\vec{d_1} = (2, 3, 4)$.

• Line 2 passes through $B(3, k, 0)$ with direction vector $\vec{d_2} = (1, 2, 1)$.
Let's compute the row elements $x_2 - x_1 = 3 - 1 = 2$, $y_2 - y_1 = k - (-1) = k + 1$, and $z_2 - z_1 = 0 - 1 = -1$. Now, substitute these terms into our coplanarity determinant equation: $$ \begin{vmatrix} 2 & k+1 & -1 \\ 2 & 3 & 4 \\ 1 & 2 & 1 \end{vmatrix} = 0 $$ Expanding the determinant along the first row: $$ 2 \cdot [3(1) - 4(2)] - (k+1) \cdot [2(1) - 4(1)] + (-1) \cdot [2(2) - 3(1)] = 0 $$ $$ 2 \cdot (3 - 8) - (k+1) \cdot (2 - 4) - 1 \cdot (4 - 3) = 0 $$ $$ 2 \cdot (-5) - (k+1) \cdot (-2) - 1 \cdot (1) = 0 $$ $$ -10 + 2(k+1) - 1 = 0 $$ $$ -11 + 2k + 2 = 0 $$ $$ 2k - 9 = 0 \implies 2k = 9 \implies k = \frac{9}{2} $$

Step 4: Final Answer:
The value of $k$ for which the lines intersect is $\frac{9}{2}$, which corresponds to option (D).