Question

Find the area of a triangle whose vertices are A(2, 2, 2), B(2, 1, 3) and C(3, 2, 1).

Hint:

Remember that the area of the parallelogram formed by vectors \( \vec{u} \) and \( \vec{v} \) is \( |\vec{u} \times \vec{v}| \). The area of the triangle with these adjacent sides is always half of the parallelogram's area. Be careful with sign conventions during the determinant calculation.

Answer 1

Step 1: Understanding the Concept:
The area of a triangle in three-dimensional space with given vertices can be found using the cross product of two vectors that represent adjacent sides of the triangle. The magnitude of this cross product is equal to the area of the parallelogram formed by these vectors, and the triangle's area is half of that.
Step 2: Key Formula or Approach:
The area of \( \triangle ABC \) is given by the formula: \[ \text{Area} = \frac{1}{2} |\vec{AB} \times \vec{AC}| \] The steps are: 1. Find the vectors for two adjacent sides, e.g., \( \vec{AB} \) and \( \vec{AC} \). 2. Compute their cross product. 3. Calculate the magnitude of the resulting vector and divide by 2.
Step 3: Detailed Explanation:
The vertices are A(2, 2, 2), B(2, 1, 3), and C(3, 2, 1).
First, determine the vectors for two sides starting from vertex A: \[ \vec{AB} = (\text{Position of B}) - (\text{Position of A}) = (2-2)\hat{i} + (1-2)\hat{j} + (3-2)\hat{k} = 0\hat{i} - 1\hat{j} + 1\hat{k} \] \[ \vec{AC} = (\text{Position of C}) - (\text{Position of A}) = (3-2)\hat{i} + (2-2)\hat{j} + (1-2)\hat{k} = 1\hat{i} + 0\hat{j} - 1\hat{k} \] Next, calculate the cross product \( \vec{AB} \times \vec{AC} \): \[ \vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
0 & -1 & 1
1 & 0 & -1 \end{vmatrix} \] \[ = \hat{i}((-1)(-1) - (1)(0)) - \hat{j}((0)(-1) - (1)(1)) + \hat{k}((0)(0) - (-1)(1)) \] \[ = \hat{i}(1 - 0) - \hat{j}(0 - 1) + \hat{k}(0 + 1) \] \[ = 1\hat{i} + 1\hat{j} + 1\hat{k} \] Now, find the magnitude of this cross product vector: \[ |\vec{AB} \times \vec{AC}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{1+1+1} = \sqrt{3} \] Finally, the area of the triangle is half of this magnitude: \[ \text{Area} = \frac{1}{2} |\vec{AB} \times \vec{AC}| = \frac{\sqrt{3}}{2} \] Step 4: Final Answer:
The area of the triangle is \( \frac{\sqrt{3}}{2} \) square units.