Question

If the vectors \( a\hat{i} + \hat{j} + 3\hat{k} \), \( 4\hat{i} + 5\hat{j} + \hat{k} \), and \( 4\hat{i} + 2\hat{j} + 6\hat{k} \) are coplanar, then \( a \) is:

• A. \( 2 \)
• B. \( 1 \)
• C. \( 3 \)
• D. \( 4 \)

Hint:

To check if three vectors are coplanar, use the scalar triple product formula: \[ \mathbf{A} \cdot (\mathbf{B} \times \mathbf{C}) = 0. \] If the determinant of the matrix formed by the three vectors is zero, they are coplanar.

Answer 1

The Correct Option is A

Step 1: Condition for Coplanarity
Three vectors are coplanar if their scalar triple product is zero: \[ \begin{vmatrix} a & 1 & 3 \\ 4 & 5 & 1 \\ 4 & 2 & 6 \end{vmatrix} = 0 \] Step 2: Compute the Determinant Expanding along the first row: \[ a \begin{vmatrix} 5 & 1 \\ 2 & 6 \end{vmatrix} - 1 \begin{vmatrix} 4 & 1 \\ 4 & 6 \end{vmatrix} + 3 \begin{vmatrix} 4 & 5 \\ 4 & 2 \end{vmatrix} = 0 \] Step 3: Evaluating the 2×2 Determinants \[ \begin{vmatrix} 5 & 1 \\ 2 & 6 \end{vmatrix} = (5)(6) - (1)(2) = 30 - 2 = 28 \] \[ \begin{vmatrix} 4 & 1 \\ 4 & 6 \end{vmatrix} = (4)(6) - (1)(4) = 24 - 4 = 20 \] \[ \begin{vmatrix} 4 & 5 \\ 4 & 2 \end{vmatrix} = (4)(2) - (5)(4) = 8 - 20 = -12 \] Step 4: Solve for \( a \) \[ a(28) - 1(20) + 3(-12) = 0 \] \[ 28a - 20 - 36 = 0 \] \[ 28a - 56 = 0 \] \[ 28a = 56 \] \[ a = 2 \] Thus, the correct answer is \( \mathbf{2} \).