Question

Find the area of the parallelogram whose diagonals are \( \vec{a} = 3\hat{i} + \hat{j} - 2\hat{k} \) and \( \vec{b} = \hat{i} - 3\hat{j} + 4\hat{k} \).

Hint:

It is crucial to distinguish between the formula for the area of a parallelogram given its adjacent sides (\(|\vec{a} \times \vec{b}|\)) and the one given its diagonals (\(\frac{1}{2}|\vec{d_1} \times \vec{d_2}|\)). A common mistake is to forget the \( \frac{1}{2} \) factor when using diagonals.

Answer 1

Step 1: Understanding the Concept:
The area of a parallelogram can be determined using the vector cross product. When the diagonals of the parallelogram are given as vectors, there is a specific formula that relates the area to the magnitude of the cross product of these diagonal vectors.
Step 2: Key Formula or Approach:
If \( \vec{d_1} \) and \( \vec{d_2} \) are the vectors representing the diagonals of a parallelogram, the area of the parallelogram is given by: \[ \text{Area} = \frac{1}{2} |\vec{d_1} \times \vec{d_2}| \] Step 3: Detailed Explanation:
Here, the diagonal vectors are \( \vec{a} = 3\hat{i} + \hat{j} - 2\hat{k} \) and \( \vec{b} = \hat{i} - 3\hat{j} + 4\hat{k} \).
First, we need to compute the cross product \( \vec{a} \times \vec{b} \). \[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
3 & 1 & -2
1 & -3 & 4 \end{vmatrix} \] \[ = \hat{i}((1)(4) - (-2)(-3)) - \hat{j}((3)(4) - (-2)(1)) + \hat{k}((3)(-3) - (1)(1)) \] \[ = \hat{i}(4 - 6) - \hat{j}(12 + 2) + \hat{k}(-9 - 1) \] \[ = -2\hat{i} - 14\hat{j} - 10\hat{k} \] Next, we find the magnitude of this resulting vector: \[ |\vec{a} \times \vec{b}| = \sqrt{(-2)^2 + (-14)^2 + (-10)^2} \] \[ = \sqrt{4 + 196 + 100} = \sqrt{300} \] \[ = \sqrt{100 \times 3} = 10\sqrt{3} \] Finally, we use the formula for the area: \[ \text{Area} = \frac{1}{2} |\vec{a} \times \vec{b}| = \frac{1}{2} (10\sqrt{3}) \] Step 4: Final Answer:
The area of the parallelogram is \( 5\sqrt{3} \) square units.