Question

If the plane $x/2 - y/3 - z/5 = 1$ cuts the co-ordinate axes in points A, B, C respectively, then the area of the triangle ABC is ______.

• A. 17/2 sq. units
• B. 19/2 sq. units
• C. 11/2 sq. units
• D. 15/2 sq. units

Hint:

There is a direct shortcut formula for the area of a triangle formed by the intercepts $(a,0,0), (0,b,0), (0,0,c)$:
$\text{Area} = \frac{1}{2} \sqrt{(ab)^2 + (bc)^2 + (ca)^2}$.
Here, $a=2, b=-3, c=-5$. Area = $\frac{1}{2} \sqrt{(-6)^2 + (15)^2 + (-10)^2} = \frac{1}{2}\sqrt{36+225+100} = \frac{\sqrt{361}}{2} = \frac{19}{2}$.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The intercepts on the $x, y,$ and $z$ axes are $a=2, b=-3, c=-5$. The points are $A(2, 0, 0), B(0, -3, 0),$ and $C(0, 0, -5)$. The area of a triangle in 3D can be found using the vector cross product.

Step 2: Formula Application:
Area $= \frac{1}{2} |\vec{AB} \times \vec{AC}|$. $\vec{AB} = (-2, -3, 0)$ and $\vec{AC} = (-2, 0, -5)$.

Step 3: Explanation:
$\vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -2 & -3 & 0 \\ -2 & 0 & -5 \end{vmatrix} = \hat{i}(15) - \hat{j}(10) + \hat{k}(-6) = 15\hat{i} - 10\hat{j} - 6\hat{k}$. Magnitude $= \sqrt{15^2 + (-10)^2 + (-6)^2} = \sqrt{225 + 100 + 36} = \sqrt{361} = 19$. Area $= \frac{1}{2} \times 19 = 19/2$.

Step 4: Final Answer:
The area is 19/2 sq. units.