<p>Given two lines <br/>\[L_1: \frac{x-1}{2} = \frac{y+1}{3} = \frac{z-1}{4}\]<br/>\[L_2: \frac{x-3}{1} = \frac{y-k}{2} = \frac{z-0}{1}\]<br/>Find the value of \(k\) such that the lines \(L_1\) and \(L_2\) are intersecting each other.</p>
Step-by-Step Solution
Key Concept: Two lines intersect if and only if there exist values of the parameters such that the corresponding points on both lines coincide, or equivalently, when the scalar triple product of the direction vector and the vector joining any two points on the lines equals zero.
Step 1: Any point $P$ on line $L_1$ is of the form $P(2p+1, 3p-1, 4p+1)$ and any point $Q$ on line $L_2$ is of the form $Q(q+3, 2q+k, q)$. Step 2: Since $L_1$ and $L_2$ are intersecting, points $P$ and $Q$ must coincide at the intersection, so corresponding coordinates must be equal: $2p+1 = q+3$
$3p-1 = 2q+k$
$4p+1 = q$ Step 3: From the first and third equations:
$2p+1 = q+3 \text{ and } 4p+1 = q$
Solving: $p = -\frac{3}{2}$ and $q = -5$ Step 4: Substitute into the second equation:
$3\left(-\frac{3}{2}\right) - 1 = 2(-5) + k$
$-\frac{9}{2} - 1 = -10 + k$
$k = \frac{9}{2}$ Alternatively, using the condition that the two lines intersect, the determinant: $\begin{vmatrix} x_1-x_2 & y_1-y_2 & z_1-z_2 \\ l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2 \end{vmatrix} = 0$ gives $k^2 + 3k = 0$, so $k = 0$ or $k = -3$. ∴ Answer is $k = 0$ or $k = -3$.
Correct Answer: 0 or -3