Question

A plane meets the co-ordinate axes in A, B, C such that the centroid of the triangle ABC is the point \((1,r,r^2)\), then the equation of the plane is,

• A. \(x+ry+r^2z = 3r^2\)
• B. \(r^2x+ry+z = 3r^2\)
• C. \(x+ry+r^2z = 3\)
• D. \(r^2x+ry+z = 3\)

Hint:

The centroid of the intercept triangle gives the three intercepts directly.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Let the plane meet the axes at \(A(a,0,0)\), \(B(0,b,0)\), \(C(0,0,c)\). The centroid of \(\triangle ABC\) is \(\left(\frac a3, \frac b3, \frac c3\right)\).

Step 2: Match the centroid:
\(\frac a3 = 1\), \(\frac b3 = r\), \(\frac c3 = r^2\), so \(a = 3\), \(b = 3r\), \(c = 3r^2\).

Step 3: Plane equation:
\[ \frac{x}{3} + \frac{y}{3r} + \frac{z}{3r^2} = 1 \]
Multiply by \(3r^2\):
\[ r^2x + ry + z = 3r^2 \]

Step 4: Why the other options are wrong.
Options (A) and (C) have \(x\) with coefficient 1 and \(z\) with \(r^2\), which is the reverse assignment of intercepts. Option (D) has the right coefficients but the wrong right-hand side \(3\) instead of \(3r^2\).

Final Answer:
The plane is \(r^2x + ry + z = 3r^2\), option (B). \[ \boxed{r^2x+ry+z=3r^2} \]