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 Question:
We are given a plane in intercept form. We must find where it intersects the $x, y$, and $z$ axes to form three vertices of a triangle in 3D space, and then calculate the geometric area of that triangle.

Step 2: Key Formula or Approach:
1. The intercept form of a plane is $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$, giving intercepts $(a,0,0), (0,b,0), (0,0,c)$.
2. The area of a triangle formed by three 3D points can be found using the cross product of two of its side vectors: $\text{Area} = \frac{1}{2} |\vec{AB} \times \vec{AC}|$.

Step 3: Detailed Explanation:
From the equation $\frac{x}{2} + \frac{y}{-3} + \frac{z}{-5} = 1$, the coordinates of the vertices are:
$A = (2, 0, 0)$
$B = (0, -3, 0)$
$C = (0, 0, -5)$
Construct the two side vectors originating from $A$:
$\vec{AB} = B - A = (0 - 2)\hat{i} + (-3 - 0)\hat{j} + (0 - 0)\hat{k} = -2\hat{i} - 3\hat{j} + 0\hat{k}$
$\vec{AC} = C - A = (0 - 2)\hat{i} + (0 - 0)\hat{j} + (-5 - 0)\hat{k} = -2\hat{i} + 0\hat{j} - 5\hat{k}$
Calculate the cross product $\vec{AB} \times \vec{AC}$:
$$\vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -2 & -3 & 0 \\ -2 & 0 & -5 \end{vmatrix}$$
$$= \hat{i}( (-3)(-5) - 0 ) - \hat{j}( (-2)(-5) - 0 ) + \hat{k}( (-2)(0) - (-3)(-2) )$$
$$= \hat{i}(15) - \hat{j}(10) + \hat{k}(0 - 6) = 15\hat{i} - 10\hat{j} - 6\hat{k}$$
Find the magnitude of this resulting cross product vector:
$$|\vec{AB} \times \vec{AC}| = \sqrt{(15)^2 + (-10)^2 + (-6)^2}$$
$$|\vec{AB} \times \vec{AC}| = \sqrt{225 + 100 + 36} = \sqrt{361} = 19$$
Finally, calculate the area of the triangle:
$$\text{Area} = \frac{1}{2} |\vec{AB} \times \vec{AC}| = \frac{19}{2}$$

Step 4: Final Answer:
The area is 19/2 sq. units, matching option (b).