Question

The area of the parallelogram with vertices \( A(1, 2, 3) \), \( B(1, 3, a) \), \( C(3, 8, 6) \) and \( D(3, 7, 3) \) is \( \sqrt{265} \) sq. units, then \( a = \)

• A. \(-5, 2\)
• B. \(6\)
• C. \(-6, 0\)
• D. \(6, 0\)

Hint:

Area of parallelogram = \(|\overrightarrow{AB} \times \overrightarrow{AD}|\). For vertices in order, take adjacent sides from a common vertex.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We have four vertices of a parallelogram. Area = \(\sqrt{265}\). Need \(a\).

Step 2: Key Formula or Approach:
In a parallelogram, \(\overrightarrow{AB}\) and \(\overrightarrow{AD}\) are adjacent sides. Area = \(|\overrightarrow{AB} \times \overrightarrow{AD}|\).

Step 3: Detailed Explanation:
\(\overrightarrow{AB} = (1-1, 3-2, a-3) = (0, 1, a-3)\).
\(\overrightarrow{AD} = (3-1, 7-2, 3-3) = (2, 5, 0)\).
Cross product: \[ \overrightarrow{AB} \times \overrightarrow{AD} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 1 & a-3 \\ 2 & 5 & 0 \end{vmatrix} = \mathbf{i}(1\cdot0 - (a-3)\cdot5) - \mathbf{j}(0\cdot0 - (a-3)\cdot2) + \mathbf{k}(0\cdot5 - 1\cdot2) \] \[ = \mathbf{i}(-5a+15) - \mathbf{j}(-2a+6) + \mathbf{k}(-2) = (-5a+15, 2a-6, -2). \] Magnitude: \[ |\overrightarrow{AB} \times \overrightarrow{AD}| = \sqrt{(-5a+15)^2 + (2a-6)^2 + (-2)^2}. \] Given area = \(\sqrt{265}\). Square both sides: \[ (5a-15)^2 + (2a-6)^2 + 4 = 265. \] \[ 25(a-3)^2 + 4(a-3)^2 + 4 = 265 \implies 29(a-3)^2 = 261 \implies (a-3)^2 = 9. \] \[ a-3 = \pm 3 \implies a = 6 \text{ or } a = 0. \] Thus \(a = 6, 0\).

Step 4: Final Answer:
Option (D) is correct.