Vector Algebra
Coplanarity of Vectors
Grade 12
Question:
<p>Let \(\alpha \in R\) and the three vectors \(\vec{a} = \alpha\hat{i} + \hat{j} + 3\hat{k}\), \(\vec{b} = 2\hat{i} + \hat{j} - \alpha\hat{k}\) and \(\vec{c} = \alpha\hat{i} - 2\hat{j} + 3\hat{k}\). Then the set \(S = \{\alpha : \vec{a}, \vec{b}\text{ and }\vec{c}\text{ are coplanar}\}\)</p>
<p>Is singleton.</p>
<p>Is empty.</p>
<p>Contains exactly two positive numbers.</p>
<p>Contains exactly two numbers only one of which is positive.</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 of the matrix formed by their components must be zero.
Step 1: Three vectors are coplanar when their scalar triple product = 0, i.e., $\vec{a} \cdot (\vec{b} \times \vec{c}) = 0$. Step 2: This is equivalent to: $\begin{vmatrix} \alpha & 1 & 3 \\ 2 & 1 & -\alpha \\ \alpha & -2 & 3 \end{vmatrix} = 0$ Step 3: 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 4: 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 5: Since $\alpha^2 = -6$ has no real solutions, the set $S = \emptyset$ (empty set). ∴ Answer: B
Correct Answer: B