Vector Algebra
Coplanarity of Vectors
Grade 12
Question:
<p>Given \(\vec{a} = \alpha\hat{i} + \hat{j} + 3\hat{k}\), \(\vec{b} = 2\hat{i} + \hat{j} - \alpha\hat{k}\) and \(\vec{c} = \alpha\hat{i} - 2\hat{j} + 3\hat{k}\) are coplanar. If \(S\) is the set of all values of \(\alpha\), then find \(S\).</p>
<p>\(S = \{\sqrt{18}\}\)</p>
<p>\(S\) is an empty set</p>
<p>\(S = \{-\sqrt{18}\}\)</p>
<p>\(S = \{\sqrt{18}, -\sqrt{18}\}\)</p>
Step-by-Step Solution
Key Concept: Three vectors are coplanar if and only if their scalar triple product equals zero, meaning the determinant of the matrix formed by their components must be zero. This reduces to finding values of α that make the determinant equal to zero.
Step 1: For coplanarity, the scalar triple product a⃗ · (b⃗ × c⃗) = 0 , which means: $\begin{vmatrix} \alpha & 1 & 3 \\ 2 & 1 & -\alpha \\ \alpha & -2 & 3 \end{vmatrix} = 0$ Step 2: Expand along the first row: $\alpha\begin{vmatrix} 1 & -\alpha \\ -2 & 3 \end{vmatrix} - 1\begin{vmatrix} 2 & -\alpha \\ \alpha & 3 \end{vmatrix} + 3\begin{vmatrix} 2 & 1 \\ \alpha & -2 \end{vmatrix} = 0$ Step 3: Calculate each 2×2 determinant: $\alpha(3 - 2\alpha) - 1(6 + \alpha^2) + 3(-4 - \alpha) = 0$ $3\alpha - 2\alpha^2 - 6 - \alpha^2 - 12 - 3\alpha = 0$ $-3\alpha^2 - 18 = 0$ $\alpha^2 = -6$ Step 4: Since α^2 = -6 has no real solutions, S = ∅ (empty set) ∴ Answer: B
Correct Answer: B