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

Step 1: Understanding the Concept:
The area of a triangle with vertices $A, B, C$ is given by $\frac{1}{2} |\vec{AB} \times \vec{AC}|$.
Step 2: Formula Application:
$\vec{AB} = (2-1)\hat{i} + (3-1)\hat{j} + (5-2)\hat{k} = \hat{i} + 2\hat{j} + 3\hat{k}$
$\vec{AC} = (1-1)\hat{i} + (5-1)\hat{j} + (5-2)\hat{k} = 0\hat{i} + 4\hat{j} + 3\hat{k}$
Cross Product $\vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
1 & 2 & 3
0 & 4 & 3 \end{vmatrix} = \hat{i}(6-12) - \hat{j}(3-0) + \hat{k}(4-0) = -6\hat{i} - 3\hat{j} + 4\hat{k}$
Step 3: Explanation:
Magnitude $|\vec{AB} \times \vec{AC}| = \sqrt{(-6)^2 + (-3)^2 + 4^2} = \sqrt{36 + 9 + 16} = \sqrt{61}$
Note: Recalculating cross product components for accuracy: $i(6-12) = -6$, $j(3-0) = 3$, $k(4-0) = 4$. Result is $\sqrt{61}$.
Area = $\frac{\sqrt{61}}{2}$.
Step 4: Final Answer:
The correct option is (d).