Question

A plane is formed by the axes whose centroid is \(\left(2, -\frac{2}{3}, \frac{1}{2}\right)\). Find the distance of the plane from the origin.

Hint:

Always simplify the intercept form to general form \(Ax + By + Cz = D\) before using the distance formula. Fractions in the square root denominator are a common source of calculation errors.

Answer 1

Step 1: Understanding the Question:
A plane intersects the axes at \( (a,0,0), (0,b,0), (0,0,c) \). The centroid of the triangle formed by these points is given. We need to find the plane's equation and then its distance from \( (0,0,0) \).
Step 2: Key Formula or Approach:
1. Centroid \( G = (\frac{a}{3}, \frac{b}{3}, \frac{c}{3}) \).
2. Plane equation: \( \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 \).
3. Distance from origin: \( p = \frac{1}{\sqrt{(1/a)^2 + (1/b)^2 + (1/c)^2}} \).
Step 3: Detailed Explanation:
Given centroid \( (2, -2/3, 1/2) \).
\( a/3 = 2 \Rightarrow a = 6 \).
\( b/3 = -2/3 \Rightarrow b = -2 \).
\( c/3 = 1/2 \Rightarrow c = 3/2 \).
Equation of plane:
\[ \frac{x}{6} + \frac{y}{-2} + \frac{z}{3/2} = 1 \Rightarrow \frac{x}{6} - \frac{y}{2} + \frac{2z}{3} = 1 \]
Multiply by 6:
\[ x - 3y + 4z - 6 = 0 \]
Distance from origin \( (0,0,0) \):
\[ d = \frac{|-6|}{\sqrt{1^2 + (-3)^2 + 4^2}} = \frac{6}{\sqrt{1+9+16}} = \frac{6}{\sqrt{26}} \]
Step 4: Final Answer:
The distance is \( 6/\sqrt{26} \).