Question

If a plane meets the coordinate axes at $A, B$ and $C$ in such a way that the centroid of triangle $ABC$ is at the point $(1,2,3)$ then the equation of the plane is

• A. $\frac{x}{1}+\frac{y}{2}+\frac{z}{3}=1$
• B. $\frac{x}{3}+\frac{y}{6}+\frac{z}{9}=1$
• C. $\frac{x}{1}+\frac{y}{2}+\frac{z}{3}=\frac{1}{3}$
• D. $\frac{x}{1}-\frac{y}{2}+\frac{z}{3}=-1$

Answer 1

The Correct Option is B

To find the equation of the plane that meets the coordinate axes at points \(A\), \(B\), and \(C\), with the centroid of triangle \(ABC\) being at the point \((1,2,3)\), we use the following approach: 

• The general form of a plane that intersects the \(x\), \(y\), and \(z\) axes at points \((a, 0, 0)\), \((0, b, 0)\), and \((0, 0, c)\) respectively is: \(\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1\).
• The coordinates of point \(A\) are \((a, 0, 0)\), of point \(B\) are \((0, b, 0)\), and of point \(C\) are \((0, 0, c)\).
• The centroid \((G)\) of triangle \(ABC\) is: \(G \left(\frac{a}{3}, \frac{b}{3}, \frac{c}{3}\right)\).
• We are given that the centroid \((G)\) is at the point \((1, 2, 3)\). So, we have:
  • \(\frac{a}{3} = 1 \Rightarrow a = 3\)
  • \(\frac{b}{3} = 2 \Rightarrow b = 6\)
  • \(\frac{c}{3} = 3 \Rightarrow c = 9\)
• Substituting these values into the plane equation \(\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1\), we get: \(\frac{x}{3} + \frac{y}{6} + \frac{z}{9} = 1\).

Therefore, the equation of the plane is \(\frac{x}{3} + \frac{y}{6} + \frac{z}{9} = 1\), which matches the second option given.