Question

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

Hint:

You can use any two adjacent vectors, for example, \( \vec{BA} \) and \( \vec{BC} \). The magnitude of their cross product will be the same, so the area will also be the same. Choose the vectors that simplify calculations (e.g., those with zero components).

Answer 1

Step 1: Understanding the Concept:
The area of a triangle formed by three points in 3D space can be calculated using the vector cross product. The area is half the magnitude of the cross product of two vectors that form adjacent sides of the triangle.
Step 2: Key Formula or Approach:
The area of \( \triangle ABC \) is given by the formula: \[ \text{Area} = \frac{1}{2} |\vec{AB} \times \vec{AC}| \] We need to: 1. Find the position vectors \( \vec{AB} \) and \( \vec{AC} \). 2. Calculate their cross product \( \vec{AB} \times \vec{AC} \). 3. Find the magnitude of the resulting vector and divide by 2.
Step 3: Detailed Explanation:
The given vertices are A(1, 1, 1), B(1, 2, 3), and C(2, 3, 1).
First, find the vectors representing two sides of the triangle: \[ \vec{AB} = \text{Position Vector of B} - \text{Position Vector of A} \] \[ \vec{AB} = (1-1)\hat{i} + (2-1)\hat{j} + (3-1)\hat{k} = 0\hat{i} + 1\hat{j} + 2\hat{k} \] \[ \vec{AC} = \text{Position Vector of C} - \text{Position Vector of A} \] \[ \vec{AC} = (2-1)\hat{i} + (3-1)\hat{j} + (1-1)\hat{k} = 1\hat{i} + 2\hat{j} + 0\hat{k} \] Next, compute the cross product \( \vec{AB} \times \vec{AC} \): \[ \vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
0 & 1 & 2
1 & 2 & 0 \end{vmatrix} \] \[ = \hat{i}((1)(0) - (2)(2)) - \hat{j}((0)(0) - (2)(1)) + \hat{k}((0)(2) - (1)(1)) \] \[ = \hat{i}(0 - 4) - \hat{j}(0 - 2) + \hat{k}(0 - 1) \] \[ = -4\hat{i} + 2\hat{j} - \hat{k} \] Now, find the magnitude of this vector: \[ |\vec{AB} \times \vec{AC}| = \sqrt{(-4)^2 + (2)^2 + (-1)^2} \] \[ = \sqrt{16 + 4 + 1} = \sqrt{21} \] Finally, the area of the triangle is half of this magnitude: \[ \text{Area of } \triangle ABC = \frac{1}{2} |\vec{AB} \times \vec{AC}| = \frac{\sqrt{21}}{2} \] Step 4: Final Answer:
The area of the triangle is \( \frac{\sqrt{21}}{2} \) square units.