Question

The plane \(\frac{x}{2}+\frac{y}{3}+\frac{z}{4} = 1\) cuts the axes at the points A, B, C then the area of triangle ABC is

• A. \(\sqrt{29}\)
• B. \(\sqrt{41}\)
• C. \(\sqrt{61}\)
• D. \(\sqrt{51}\)

Hint:

Find the three axis intercept points and use half the magnitude of the cross product.

Answer 1

The Correct Option is C

Step 1: Intercept Points:
The plane meets the axes at \(A(2,0,0)\), \(B(0,3,0)\) and \(C(0,0,4)\).

Step 2: Edge Vectors:
\(\overrightarrow{AB}=(-2,3,0)\) and \(\overrightarrow{AC}=(-2,0,4)\).

Step 3: Cross Product:
\[ \overrightarrow{AB}\times\overrightarrow{AC}=(3\cdot4-0\cdot0,\ 0\cdot(-2)-(-2)(4),\ (-2)(0)-3(-2))=(12,8,6) \]

Step 4: Area:
\[ \text{Area}=\frac12\sqrt{144+64+36}=\frac12\sqrt{244}=\frac12\cdot2\sqrt{61}=\sqrt{61} \]
Option (C) is correct.

Final Answer:
The area is \(\sqrt{61}\), option (C). \[ \boxed{\text{(C) } \sqrt{61}} \]