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

The general equation of the plane meeting the coordinate axes at \( A(x_1, 0, 0) \), \( B(0, y_1, 0) \), and \( C(0, 0, z_1) \) is: \[ \frac{x}{x_1} + \frac{y}{y_1} + \frac{z}{z_1} = 1 \] Now, the centroid of the triangle formed by points \( A \), \( B \), and \( C \) is given by: \[ \text{Centroid} = \left( \frac{x_1}{3}, \frac{y_1}{3}, \frac{z_1}{3} \right) \] We are told that the centroid is at the point \( (1, 2, 3) \). So: \[ \frac{x_1}{3} = 1, \quad \frac{y_1}{3} = 2, \quad \frac{z_1}{3} = 3 \] From this, we get: \[ x_1 = 3, \quad y_1 = 6, \quad z_1 = 9 \] Substituting these values into the general equation of the plane: \[ \frac{x}{3} + \frac{y}{6} + \frac{z}{9} = 1 \]

So, the correct answer is (B) : \(\frac{x}{3}+\frac{y}{6}+\frac{z}{9}=1\).

Answer 2

Given: A plane meets coordinate axes at points A, B, and C such that the triangle ABC has centroid at point (1, 2, 3).

Let the intercepts of the plane be at: 

A = (a, 0, 0), B = (0, b, 0), C = (0, 0, c)

Then, the equation of the plane is of the form:

\[ \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 \]

Centroid of triangle ABC is:

\[ G = \left(\frac{a+0+0}{3}, \frac{0+b+0}{3}, \frac{0+0+c}{3}\right) = \left(\frac{a}{3}, \frac{b}{3}, \frac{c}{3}\right) \] Given: \( G = (1, 2, 3) \)

So,

\[ \frac{a}{3} = 1 \Rightarrow a = 3 \\ \frac{b}{3} = 2 \Rightarrow b = 6 \\ \frac{c}{3} = 3 \Rightarrow c = 9 \]

Therefore, the equation of the plane is:

\[ \frac{x}{3} + \frac{y}{6} + \frac{z}{9} = 1 \]

Final Answer: \( \boxed{\frac{x}{3} + \frac{y}{6} + \frac{z}{9} = 1} \)