Question

The plane $\frac{x}{2} + \frac{y}{3} + \frac{z}{4} = 1$ cuts the $X$-axis at $A$, $Y$-axis at $B$ and $Z$-axis at $C$, then the area of $\Delta ABC$ is

• A. $\sqrt{61}$ sq. units
• B. $29$ sq. units
• C. $61$ sq. units
• D. $\sqrt{29}$ sq. units

Hint:

For any triangle formed by the axes-intercepts $a$, $b$, and $c$ of a plane, the area can be directly evaluated using the neat three-dimensional Pythagorean area shortcut:
$$\text{Area} = \frac{1}{2}\sqrt{(ab)^2 + (bc)^2 + (ca)^2}$$ Plugging in $a=2, b=3, c=4$:
$$\text{Area} = \frac{1}{2}\sqrt{(6)^2 + (12)^2 + (8)^2} = \frac{1}{2}\sqrt{36 + 144 + 64} = \frac{1}{2}\sqrt{244} = \sqrt{61}\ \text{sq. units.}$$

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
A plane x/2 + y/3 + z/4 = 1 meets axes at A, B, C. Find the area of triangle ABC.

Step 2: Key Formula or Approach:
Vertices are A(a,0,0), B(0,b,0), C(0,0,c). Form vectors AB⃗ and AC⃗, then Area = ½|AB⃗ × AC⃗|.

Step 3: Detailed Explanation:
Intercepts: a=2, b=3, c=4. Vertices: A(2,0,0), B(0,3,0), C(0,0,4). AB⃗ = -2î + 3ĵ, AC⃗ = -2î + 4k̂. Cross product: |î ĵ k̂; -2 3 0; -2 0 4| = î(12) - ĵ(-8) + k̂(6) = 12î + 8ĵ + 6k̂. Magnitude = √(144+64+36) = √244 = 2√61. Area = ½(2√61) = √61 sq. units.

Step 4: Final Answer:
The area is √61 sq. units, option (A).