Question

Find the area of a parallelogram whose adjacent sides are the vectors \( \vec{a} = \hat{i} - \hat{j} + 3\hat{k} \) and \( \vec{b} = 2\hat{i} - 7\hat{j} + \hat{k} \).

Hint:

When calculating determinants for cross products, be very careful with the signs, especially for the \( \hat{j} \) component which is preceded by a minus sign. A small sign error can lead to a completely different result.

Answer 1

Step 1: Understanding the Concept:
The area of a parallelogram formed by two adjacent vectors \( \vec{a} \) and \( \vec{b} \) is given by the magnitude of their cross product, i.e., Area = \( |\vec{a} \times \vec{b}| \).
Step 2: Key Formula or Approach:
1. Compute the cross product \( \vec{a} \times \vec{b} \).
2. Calculate the magnitude of the resulting vector.
The cross product is calculated using the determinant of a matrix: \[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
a_x & a_y & a_z
b_x & b_y & b_z \end{vmatrix} \] Step 3: Detailed Explanation:
Given vectors: \[ \vec{a} = 1\hat{i} - 1\hat{j} + 3\hat{k} \] \[ \vec{b} = 2\hat{i} - 7\hat{j} + 1\hat{k} \] First, we find the cross product \( \vec{a} \times \vec{b} \): \[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
1 & -1 & 3
2 & -7 & 1 \end{vmatrix} \] \[ = \hat{i}((-1)(1) - (3)(-7)) - \hat{j}((1)(1) - (3)(2)) + \hat{k}((1)(-7) - (-1)(2)) \] \[ = \hat{i}(-1 + 21) - \hat{j}(1 - 6) + \hat{k}(-7 + 2) \] \[ = 20\hat{i} + 5\hat{j} - 5\hat{k} \] Next, we find the magnitude of this new vector: \[ |\vec{a} \times \vec{b}| = \sqrt{(20)^2 + (5)^2 + (-5)^2} \] \[ = \sqrt{400 + 25 + 25} \] \[ = \sqrt{450} = \sqrt{225 \times 2} = 15\sqrt{2} \] Step 4: Final Answer:
The area of the parallelogram is \( 15\sqrt{2} \) square units.