Question

Find the area of a parallelogram whose adjacent sides are given by vectors \( \vec{a} = 3\hat{i} + \hat{j} + 2\hat{k} \) and \( \vec{b} = \hat{i} + 2\hat{j} - 2\hat{k} \).

Hint:

Remember the geometric interpretation of the cross product: its magnitude gives the area of the parallelogram formed by the two vectors. Be careful with signs when calculating the determinant for the cross product.

Answer 1

Step 1: Understanding the Concept:
The area of a parallelogram whose adjacent sides are represented by two vectors is equal to the magnitude of the cross product of those two vectors.
Step 2: Key Formula or Approach:
Area of parallelogram = \( |\vec{a} \times \vec{b}| \).
The cross product is calculated using the determinant:
\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
a_1 & a_2 & a_3
b_1 & b_2 & b_3 \end{vmatrix} \] Step 3: Detailed Explanation or Calculation:
The given vectors are \( \vec{a} = 3\hat{i} + \hat{j} + 2\hat{k} \) and \( \vec{b} = \hat{i} + 2\hat{j} - 2\hat{k} \).
First, we compute the cross product \( \vec{a} \times \vec{b} \):
\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
3 & 1 & 2
1 & 2 & -2 \end{vmatrix} \] Expanding the determinant:
\[ = \hat{i}((1)(-2) - (2)(2)) - \hat{j}((3)(-2) - (2)(1)) + \hat{k}((3)(2) - (1)(1)) \] \[ = \hat{i}(-2 - 4) - \hat{j}(-6 - 2) + \hat{k}(6 - 1) \] \[ = \hat{i}(-6) - \hat{j}(-8) + \hat{k}(5) \] \[ = -6\hat{i} + 8\hat{j} + 5\hat{k} \] Next, we find the magnitude of this resulting vector:
\[ |\vec{a} \times \vec{b}| = \sqrt{(-6)^2 + (8)^2 + (5)^2} \] \[ = \sqrt{36 + 64 + 25} \] \[ = \sqrt{125} = \sqrt{25 \times 5} = 5\sqrt{5} \] Step 4: Final Answer:
The area of the parallelogram is \( 5\sqrt{5} \) square units.