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:
The given equation of the plane is in intercept form, $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$. The plane intersects the coordinate axes at points $A$, $B$, and $C$. We need to find the area of the triangle formed by these three points of intersection in 3D space.

Step 2: Key Formula or Approach:
1. Identify the coordinates of the vertices: $A(a,0,0)$, $B(0,b,0)$, and $C(0,0,c)$. 2. Form the vectors $$\vec{AB}=-a\hat{i}+b\hat{j}+0\hat{k}, \qquad \vec{AC}=-a\hat{i}+0\hat{j}+c\hat{k}.$$ 3. The area of $\Delta ABC$ is given by $$\text{Area}=\frac{1}{2}\left|\vec{AB}\times\vec{AC}\right|.$$

Step 3: Detailed Explanation:
From the plane equation $\frac{x}{2}+\frac{y}{3}+\frac{z}{4}=1$, the intercepts are $a=2$, $b=3$, and $c=4$. Therefore, the vertices are $A(2,0,0)$, $B(0,3,0)$, and $C(0,0,4)$. The vectors are $$\vec{AB}=(0-2)\hat{i}+(3-0)\hat{j}+(0-0)\hat{k}=-2\hat{i}+3\hat{j},$$ $$\vec{AC}=(0-2)\hat{i}+(0-0)\hat{j}+(4-0)\hat{k}=-2\hat{i}+4\hat{k}.$$ Now, $$\vec{AB}\times\vec{AC}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\-2&3&0\\-2&0&4\end{vmatrix}=12\hat{i}+8\hat{j}+6\hat{k}.$$ Therefore, $$\left|\vec{AB}\times\vec{AC}\right|=\sqrt{12^2+8^2+6^2}=\sqrt{144+64+36}=\sqrt{244}=2\sqrt{61}.$$ Hence, $$\text{Area}=\frac{1}{2}\left(2\sqrt{61}\right)=\sqrt{61}\ \text{sq. units}.$$

Step 4: Final Answer:
The area of the triangle is $\sqrt{61}$ sq. units, which corresponds to option (A).