Question

Find the area of a triangle with vertices \( A(1,2,3), B(2,3,4), \) and \( C(3,4,5) \).

Hint:

If two side vectors are scalar multiples, the triangle is degenerate and its area is zero.

Answer 1

Concept: The area of a triangle formed by three points in 3D space is given by: \[ \text{Area} = \frac{1}{2} \left| \vec{AB} \times \vec{AC} \right| \] where:

\( \vec{AB} = B - A \)
\( \vec{AC} = C - A \)
The magnitude of the cross product gives the area of the parallelogram.

Step 1: Find vectors \[ \vec{AB} = (2-1, 3-2, 4-3) = (1,1,1) \] \[ \vec{AC} = (3-1, 4-2, 5-3) = (2,2,2) \]
Step 2: Compute cross product \[ \vec{AB} \times \vec{AC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k}
1 & 1 & 1
2 & 2 & 2 \end{vmatrix} \] Since the rows are proportional, the determinant is: \[ \vec{AB} \times \vec{AC} = (0,0,0) \]
Step 3: Find area \[ \text{Area} = \frac{1}{2} \times 0 = 0 \] Final Answer: \[ \boxed{0} \] Explanation: The vectors \( \vec{AB} \) and \( \vec{AC} \) are parallel, meaning the three points are collinear. Hence, they do not form a triangle, and the area is zero.