Question

Let \(A(2,3,0)\), \(B(0,3,2)\) and \(C(4,0,3)\) be vertices of a triangle, then the area of the triangle is

• A. \(\frac{\sqrt{171}}{2}\text{ sq. units}\)
• B. \(\frac{\sqrt{172}}{2}\text{ sq. units}\)
• C. \(\frac{\sqrt{173}}{2}\text{ sq. units}\)
• D. \(\frac{\sqrt{174}}{2}\text{ sq. units}\)

Hint:

Area $=\frac12|\vec{AB}\times\vec{AC}|$.

Answer 1

The Correct Option is B

Step 1: Edge vectors
\(\vec{AB}=(-2,0,2)\) and \(\vec{AC}=(2,-3,3)\).

Step 2: Cross product
\[ \vec{AB}\times\vec{AC}=(0\cdot3-2\cdot(-3))\hat i-((-2)(3)-2\cdot2)\hat j+((-2)(-3)-0)\hat k=6\hat i+10\hat j+6\hat k \]

Step 3: Area
\(|\ldots|=\sqrt{36+100+36}=\sqrt{172}\), so the area is \(\frac{\sqrt{172}}{2}\). Option (B).

Final Answer:
The area is \(\frac{\sqrt{172}}{2}\) square units, option (B). \[ \boxed{\text{(B)}} \]