Question

Using vectors, find the area of \( \Delta ABC \) with vertices \( A(1, 2, 3) \), \( B(2, -1, 4) \) and \( C(4, 5, -1) \).

Hint:

You can use any two adjacent sides (e.g., \( \vec{BA} \) and \( \vec{BC} \)) to get the same area.
Double-check your determinant calculation; cross-products are common sources of small errors.

Answer 1

Concept:
• Vector area of a triangle: The area of a triangle with vertices \( A, B, C \) is given by: \[ \text{Area} = \frac{1}{2} | \vec{AB} \times \vec{AC} | \]
• Position vectors: \( \vec{AB} = \text{P.V. of B} - \text{P.V. of A} \).

Step 1: Determine the vectors \( \vec{AB} \) and \( \vec{AC} \)
\[ \vec{AB} = (2 - 1)\hat{i} + (-1 - 2)\hat{j} + (4 - 3)\hat{k} = \hat{i} - 3\hat{j} + \hat{k} \] \[ \vec{AC} = (4 - 1)\hat{i} + (5 - 2)\hat{j} + (-1 - 3)\hat{k} = 3\hat{i} + 3\hat{j} - 4\hat{k} \]

Step 2: Calculate the cross product \( \vec{AB} \times \vec{AC} \)
\[ \vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & 1 \\ 3 & 3 & -4 \end{vmatrix} \] \[ = \hat{i} [(-3)(-4) - (3)(1)] - \hat{j} [(1)(-4) - (3)(1)] + \hat{k} [(1)(3) - (3)(-3)] \] \[ = \hat{i} [12 - 3] - \hat{j} [-4 - 3] + \hat{k} [3 + 9] \] \[ = 9\hat{i} + 7\hat{j} + 12\hat{k} \]

Step 3: Find the magnitude and calculate area
\[ | \vec{AB} \times \vec{AC} | = \sqrt{9^2 + 7^2 + 12^2} \] \[ = \sqrt{81 + 49 + 144} = \sqrt{274} \] \[ \text{Area} = \frac{1}{2} \sqrt{274} \text{ sq. units} \]