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 Question:
We are given a plane in intercept form. We must find the coordinates where it cuts the axes and compute the area of the 3D triangle formed by these three intercept points.

Step 2: Detailed Explanation:
The equation of the plane is strictly in intercept form: $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$.
From the equation, the intercepts are:
$a = 2$, so point A is $(2, 0, 0)$
$b = 3$, so point B is $(0, 3, 0)$
$c = 6$, so point C is $(0, 0, 6)$
The formula for the area of a triangle formed by the intercepts $(a,0,0), (0,b,0), (0,0,c)$ on the coordinate axes is derived using vector cross products and evaluates to:
$\text{Area} = \frac{1}{2} \sqrt{(ab)^2 + (bc)^2 + (ca)^2}$
Substitute the values $a=2, b=3, c=6$:
$ab = 2 \times 3 = 6$
$bc = 3 \times 6 = 18$
$ca = 6 \times 2 = 12$
Now, plug these into the area formula:
$\text{Area} = \frac{1}{2} \sqrt{(6)^2 + (18)^2 + (12)^2}$
$\text{Area} = \frac{1}{2} \sqrt{36 + 324 + 144}$
$\text{Area} = \frac{1}{2} \sqrt{504}$
Simplify the radical by finding a large perfect square factor. Notice $504 = 36 \times 14$:
$\text{Area} = \frac{1}{2} \sqrt{36 \times 14}$
$\text{Area} = \frac{1}{2} \times 6\sqrt{14}$
$\text{Area} = 3\sqrt{14} \text{ sq. units}$.

Step 3: Final Answer:
The area is $3\sqrt{14}$ sq. units, matching option (b).