Question

If \( \mathbf{a} = \mathbf{i} + \mathbf{j} + \mathbf{k}, \mathbf{b} = \mathbf{i} + 3\mathbf{j} + 5\mathbf{k}, \) and \( \mathbf{c} = 7\mathbf{i} + 9\mathbf{j} + 11\mathbf{k} \), then the area of the parallelogram having diagonals \( \mathbf{a} + \mathbf{b} \) and \( \mathbf{b} + \mathbf{c} \) is:

• A. \( 4\sqrt{6} \) sq. units
• B. \( \frac{1}{2} \sqrt{21} \) sq. units
• C. \( \frac{\sqrt{6}}{2} \) sq. units
• D. \( \sqrt{6} \) sq. units

Hint:

To find the area of a parallelogram with given diagonals, use the cross product of the vectors corresponding to the diagonals and compute its magnitude.

Answer 1

The Correct Option is A

Step 1: Understanding the problem.
We are given the vectors \( \mathbf{a}, \mathbf{b}, \mathbf{c} \), and we need to find the area of a parallelogram formed by the diagonals \( \mathbf{a} + \mathbf{b} \) and \( \mathbf{b} + \mathbf{c} \). The formula for the area of a parallelogram formed by two vectors \( \mathbf{u} \) and \( \mathbf{v} \) is given by: \[ \text{Area} = \left| \mathbf{u} \times \mathbf{v} \right|. \] Here, the vectors are \( \mathbf{u} = \mathbf{a} + \mathbf{b} \) and \( \mathbf{v} = \mathbf{b} + \mathbf{c} \).
Step 2: Calculating the cross product.
First, find the vectors \( \mathbf{u} \) and \( \mathbf{v} \): \[ \mathbf{u} = \mathbf{a} + \mathbf{b} = (1+1)\mathbf{i} + (1+3)\mathbf{j} + (1+5)\mathbf{k} = 2\mathbf{i} + 4\mathbf{j} + 6\mathbf{k}, \] \[ \mathbf{v} = \mathbf{b} + \mathbf{c} = (1+7)\mathbf{i} + (3+9)\mathbf{j} + (5+11)\mathbf{k} = 8\mathbf{i} + 12\mathbf{j} + 16\mathbf{k}. \] Now, compute the cross product \( \mathbf{u} \times \mathbf{v} \): \[ \mathbf{u} \times \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 4 & 6 \\ 8 & 12 & 16 \end{vmatrix}. \] Expanding the determinant: \[ \mathbf{u} \times \mathbf{v} = \mathbf{i} \begin{vmatrix} 4 & 6 \\ 12 & 16 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 2 & 6 \\ 8 & 16 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 2 & 4 \\ 8 & 12 \end{vmatrix}. \] \[ \mathbf{u} \times \mathbf{v} = \mathbf{i} \left( 4(16) - 6(12) \right) - \mathbf{j} \left( 2(16) - 6(8) \right) + \mathbf{k} \left( 2(12) - 4(8) \right), \] \[ \mathbf{u} \times \mathbf{v} = \mathbf{i} (64 - 72) - \mathbf{j} (32 - 48) + \mathbf{k} (24 - 32), \] \[ \mathbf{u} \times \mathbf{v} = -8\mathbf{i} + 16\mathbf{j} - 8\mathbf{k}. \]
Step 3: Finding the magnitude of the cross product.
Now, calculate the magnitude of \( \mathbf{u} \times \mathbf{v} \): \[ \left| \mathbf{u} \times \mathbf{v} \right| = \sqrt{(-8)^2 + 16^2 + (-8)^2} = \sqrt{64 + 256 + 64} = \sqrt{384} = 4\sqrt{6}. \]
Step 4: Conclusion.
Therefore, the area of the parallelogram is \( 4\sqrt{6} \) square units, and the correct answer is (a).