3D Geometry
Coplanarity of lines
Grade 12
Question:
<p>Given lines are: \(\vec{r} = (\hat{i} + (-3\hat{j}) + \hat{k}) + s(\hat{i} - \lambda\hat{j} + \lambda\hat{k})\) and \(\vec{r} = (0\hat{i} + \hat{j} + 2\hat{k}) + t\left(\dfrac{1}{2}\hat{i} + \hat{j} - \hat{k}\right)\). If \(\vec{a} = (1-0)\hat{i} + (-3-1)\hat{j} + (1-2)\hat{k}\), \(\vec{b} = \hat{i} - \lambda\hat{j} + \lambda\hat{k}\), and \(\vec{c} = \dfrac{1}{2}\hat{i} + \hat{j} - \hat{k}\) are coplanar, then \(\lambda\) equals:</p>
<p>(A) \(\lambda = 0\)</p>
<p>(B) \(\lambda = -2\)</p>
<p>(C) \(\lambda = 2\)</p>
<p>(D) \(\lambda = 1\)</p>
Step-by-Step Solution
Key Concept: Three vectors are coplanar if and only if their scalar triple product equals zero, which means the determinant formed by these vectors must equal zero.
Step 1: Identify the vectors. a = (1, -4, -1), b = (1, -λ, λ), c = (1/2, 1, -1) Step 2: For coplanarity, the scalar triple product must be zero: $\begin{vmatrix} 1 & -4 & -1 \\ 1 & -\lambda & \lambda \\ 1/2 & 1 & -1 \end{vmatrix} = 0$ Step 3: Expand along the first row: $1 \begin{vmatrix} -\lambda & \lambda \\ 1 & -1 \end{vmatrix} - (-4) \begin{vmatrix} 1 & \lambda \\ 1/2 & -1 \end{vmatrix} + (-1) \begin{vmatrix} 1 & -\lambda \\ 1/2 & 1 \end{vmatrix} = 0$ Step 4: Calculate 2×2 determinants: $1(\lambda - \lambda) + 4(-1 - \lambda/2) - (1 + \lambda/2) = 0$ $0 - 4 - 2\lambda - 1 - \lambda/2 = 0$ $-5 - \frac{5\lambda}{2} = 0$ $\lambda = -2$ ∴ Answer: B (λ = -2)
Correct Answer: B