Question

If the vectors \( 2\vec{i} - \vec{j} + 3\vec{k}, \vec{i} + 4\vec{j} + \vec{k}, 4\vec{i} + p\vec{j} + \vec{k} \) are coplanar, then \( p = \)

• A. \( 53 \)
• B. \( 37 \)
• C. \( 43 \)
• D. \( 59 \)

Hint:

The scalar triple product \( \vec{A} \cdot (\vec{B} \times \vec{C}) \) helps to determine coplanarity. If the result is zero, the vectors are coplanar. When solving determinant-based problems, expand along rows or columns that simplify calculations.

Answer 1

The Correct Option is C

To determine the value of \( p \), we check for coplanarity by ensuring that the scalar triple product of the three vectors is zero. Let the given vectors be: \[ \vec{A} = 2\vec{i} - \vec{j} + 3\vec{k} \] \[ \vec{B} = \vec{i} + 4\vec{j} + \vec{k} \] \[ \vec{C} = 4\vec{i} + p\vec{j} + \vec{k} \] Calculate the scalar triple product: \[ \vec{A} \cdot (\vec{B} \times \vec{C}) = 0 \] Find \( \vec{B} \times \vec{C} \) using determinant form: \[ \vec{B} \times \vec{C} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 1 & 4 & 1 \\ 4 & p & 1 \end{vmatrix} \] Expanding the determinant: \[ \vec{i} (4 \cdot 1 - p \cdot 1) - \vec{j} (1 \cdot 1 - 4 \cdot 1) + \vec{k} (1 \cdot p - 4 \cdot 4) \] \[ \vec{i} (4 - p) - \vec{j} (1 - 4) + \vec{k} (p - 16) \] \[ \vec{i} (4 - p) + \vec{j} (3) + \vec{k} (p - 16) \] Now, compute \( \vec{A} \cdot (\vec{B} \times \vec{C}) \): \[ (2\vec{i} - \vec{j} + 3\vec{k}) \cdot ((4 - p) \vec{i} + 3 \vec{j} + (p - 16) \vec{k}) = 0 \] \[ 2(4 - p) - (3) + 3(p - 16) = 0 \] \[ 8 - 2p - 3 + 3p - 48 = 0 \] \[ -43 + p = 0 \] \[ p = 43 \] Thus, the correct answer is \( p = 43 \).