<p><strong>Ex. 46</strong> Two lines whose equations are \(\frac{x}{-2} = \frac{y}{-3} = \frac{z}{-2}\) and \(\frac{x-3}{2} = \frac{y-2}{3} = \frac{z-1}{\lambda}\) lie in the same plane.</p><p>The value of \(\sin^{-1} \sin \lambda\) is equal to</p>
Step-by-Step Solution
Key Concept: Two lines are coplanar if the scalar triple product of (point difference) and (direction cross product) equals zero. For inverse sine, if the argument is outside \([-\pi/2, \pi/2]\), use the identity \(\sin^{-1}(\sin \theta) = \pi - \theta\) for \(\theta \in (\pi/2, \pi)\).
Solution: For two lines to be coplanar, the scalar triple product condition must be satisfied. Line 1: passes through origin with direction \(\mathbf{d_1} = (-2, -3, -2)\) Line 2: passes through \((3, 2, 1)\) with direction \(\mathbf{d_2} = (2, 3, \lambda)\) Coplanarity condition: \(\mathbf{(P_2 - P_1) \cdot (d_1 \times d_2)} = 0\) \((3, 2, 1) \cdot \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -2 & -3 & -2 \\ 2 & 3 & \lambda \end{vmatrix} = 0\) \(= (3, 2, 1) \cdot (-3\lambda + 6, 2\lambda + 4, -6 + 6)\) \(= 3(-3\lambda + 6) + 2(2\lambda + 4) + 0 = 0\) \(-9\lambda + 18 + 4\lambda + 8 = 0\) \(-5\lambda + 26 = 0 \Rightarrow \lambda = \frac{26}{5}\) Since \(\lambda = 3 > \pi/2\), we have \(\sin^{-1}(\sin 3) = \pi - 3\)
Correct Answer: B