Question

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

• A. \(\sqrt{61}\) sq. units
• B. \(\frac{\sqrt{61}}{2}\) sq. units
• C. \(\frac{\sqrt{61}}{4}\) sq. units
• D. \(\frac{\sqrt{71}}{2}\) sq. units

Hint:

Find the intercept points A, B and C, then compute half the magnitude of the cross product of two sides.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The plane meets the axes at \(A(2, 0, 0)\), \(B(0, 3, 0)\) and \(C(0, 0, 4)\). The area of triangle ABC is \(\dfrac{1}{2}|\overrightarrow{AB}\times\overrightarrow{AC}|\).

Step 2: Find the sides.
\(\overrightarrow{AB} = (-2, 3, 0)\) and \(\overrightarrow{AC} = (-2, 0, 4)\).

Step 3: Cross product.
\[ \overrightarrow{AB}\times\overrightarrow{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -2 & 3 & 0 \\ -2 & 0 & 4 \end{vmatrix} = (12, 8, 6) \]

Step 4: Area.
\[ \text{Area} = \frac{1}{2}\sqrt{144 + 64 + 36} = \frac{1}{2}\sqrt{244} = \frac{1}{2}\cdot 2\sqrt{61} = \sqrt{61} \]

Step 5: Check the options.
Option (A) matches. Option (B) is half of it, and option (C) is a quarter of it.

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