Question

Let \( \vec{a} = \hat{i} + \hat{j} + \hat{k}, \, \vec{b} = \hat{i} + 3\hat{j} + 5\hat{k} \) and \( \vec{c} = 7\hat{i} + 9\hat{j} + 11\hat{k} \). Then the area of a parallelogram having diagonals \( \vec{a} + \vec{b} \) and \( \vec{b} + \vec{c} \) is:

• A. \( 4\sqrt{6} \, \text{sq units} \)
• B. \( 4\sqrt{6} \, \text{sq units} \)
• C. \( \sqrt{6} \, \text{sq units} \)
• D. \( 6\sqrt{6} \, \text{sq units} \)

Hint:

Use the cross product to find areas of parallelograms formed by vectors. It simplifies the calculation of magnitude and provides the desired result.

Answer 1

The Correct Option is A

Given vectors: \[ \vec{a} = \hat{i} + \hat{j} + \hat{k}, \quad \vec{b} = \hat{i} + 3\hat{j} + 5\hat{k}, \quad \vec{c} = 7\hat{i} + 9\hat{j} + 11\hat{k}. \] Step 1: Compute the diagonal vectors. The diagonals of the parallelogram are the sums of adjacent vectors: \[ \vec{d}_1 = \vec{a} + \vec{b} = (1+1)\hat{i} + (1+3)\hat{j} + (1+5)\hat{k} = 2\hat{i} + 4\hat{j} + 6\hat{k}, \] \[ \vec{d}_2 = \vec{b} + \vec{c} = (1+7)\hat{i} + (3+9)\hat{j} + (5+11)\hat{k} = 8\hat{i} + 12\hat{j} + 16\hat{k}. \] Step 2: Calculate the cross product \( \vec{d}_1 \times \vec{d}_2 \). Using the determinant of a 3x3 matrix, we get: \[ \vec{d}_1 \times \vec{d}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
2 & 4 & 6
8 & 12 & 16 \end{vmatrix}. \] Expanding the determinant: \[ \vec{d}_1 \times \vec{d}_2 = \hat{i} \left( 4 \times 16 - 6 \times 12 \right) - \hat{j} \left( 2 \times 16 - 6 \times 8 \right) + \hat{k} \left( 2 \times 12 - 4 \times 8 \right). \] Simplifying: \[ = \hat{i} (64 - 72) - \hat{j} (32 - 48) + \hat{k} (24 - 32) \] \[ = -8\hat{i} + 16\hat{j} - 8\hat{k}. \] Step 3: Calculate the magnitude of the cross product. The magnitude is: \[ |\vec{d}_1 \times \vec{d}_2| = \sqrt{(-8)^2 + 16^2 + (-8)^2} = \sqrt{64 + 256 + 64} = \sqrt{384} = 8\sqrt{6}. \] Step 4: Calculate the area of the parallelogram. The area is half the magnitude of the cross product: \[ \text{Area} = \frac{1}{2} |\vec{d}_1 \times \vec{d}_2| = 4\sqrt{6}. \] Final Answer: \[ \boxed{4\sqrt{6} \, \text{sq units}} \]