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 Concept:
A plane intersecting the axes at \((a, 0, 0)\), \((0, b, 0)\), and \((0, 0, c)\) has a centroid given by \((\frac{a}{3}, \frac{b}{3}, \frac{c}{3})\). Once \(a, b, c\) (intercepts) are found, we write the equation of the plane.

Step 2: Key Formula or Approach:
1. Intercept form of plane: \(\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1\).
2. Distance from origin to plane \(Ax + By + Cz + D = 0\) is \(d = \frac{|D|}{\sqrt{A^2 + B^2 + C^2}}\).

Step 3: Detailed Explanation:
Let the intercepts be \(a, b, c\). The centroid is:
\[ \left(\frac{a}{3}, \frac{b}{3}, \frac{c}{3}\right) = \left(2, -\frac{2}{3}, \frac{1}{2}\right) \] Thus, \(a = 6, b = -2, c = \frac{3}{2}\).
The equation of the plane is:
\[ \frac{x}{6} + \frac{y}{-2} + \frac{z}{3/2} = 1 \] \[ \frac{x}{6} - \frac{y}{2} + \frac{2z}{3} = 1 \] Multiplying by 6 to clear fractions:
\[ x - 3y + 4z = 6 \implies x - 3y + 4z - 6 = 0 \] 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 \(\frac{6}{\sqrt{26}}\).