Question

If the plane $\frac{x}{2} + \frac{y}{3} + \frac{z}{6} = 1$ cuts the co-ordinate axes at points A, B, C respectively, then area of the triangle ABC is ______.

• A. $\sqrt{14}$ sq. units
• B. $3\sqrt{14}$ sq. units
• C. $\frac{1}{\sqrt{14}}$ sq. units
• D. $3\sqrt{13}$ sq. units

Hint:

The expression $\frac{1}{2} \sqrt{(ab)^2 + (bc)^2 + (ca)^2}$ is derived from the Pythagorean theorem in 3D areas. The total area vector is the vector sum of the three projected 2D triangle areas on the $xy, yz,$ and $zx$ planes!

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The intercepts of the plane on the $x, y, z$ axes are $a, b, c$. The coordinates of the vertices are $A(a,0,0)$, $B(0,b,0)$, and $C(0,0,c)$.

Step 2: Formula Application:
Area of $\triangle ABC = \frac{1}{2} \sqrt{(ab)^2 + (bc)^2 + (ca)^2}$.

Step 3: Explanation:
From the equation, $a=2, b=3, c=6$. $ab = 6$, $bc = 18$, $ca = 12$. Area $= \frac{1}{2} \sqrt{6^2 + 18^2 + 12^2} = \frac{1}{2} \sqrt{36 + 324 + 144}$ Area $= \frac{1}{2} \sqrt{504} = \frac{1}{2} \sqrt{36 \times 14} = \frac{1}{2} \times 6\sqrt{14} = 3\sqrt{14}$.

Step 4: Final Answer:
The area is $3\sqrt{14}$ sq. units.