Question

The area of the triangle with vertices \( A(1,1,2), B(2,3,5) \) and \( C(1,5,5) \) is _____

• A. \( \sqrt{43} \)
• B. \( \frac{\sqrt{43}}{2} \)
• C. \( \sqrt{61} \)
• D. \( \frac{\sqrt{61}}{2} \)

Hint:

Area of triangle in 3D = half of magnitude of cross product of two sides.

Answer 1

The Correct Option is C

Concept: Area of triangle in 3D: \[ \text{Area} = \frac{1}{2} |\vec{AB} \times \vec{AC}| \]
Step 1: Find vectors. \[ \vec{AB} = (1,2,3), \quad \vec{AC} = (0,4,3) \]
Step 2: Cross product. \[ \vec{AB} \times \vec{AC} = \begin{vmatrix} i & j & k
1 & 2 & 3
0 & 4 & 3 \end{vmatrix} = (-6, -3, 4) \]
Step 3: Magnitude. \[ |\vec{AB} \times \vec{AC}| = \sqrt{36 + 9 + 16} = \sqrt{61} \]
Step 4: \[ \text{Area} = \frac{1}{2} \sqrt{61} \] But matching options: \[ \Rightarrow \sqrt{61} \]