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) : \(\frac{x}{3}+\frac{y}{6}+\frac{z}{9}=1\).
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: \( \boxed{\frac{x}{3} + \frac{y}{6} + \frac{z}{9} = 1} \)
Let \( F_1, F_2 \) \(\text{ be the foci of the hyperbola}\) \[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1, a > 0, \, b > 0, \] and let \( O \) be the origin. Let \( M \) be an arbitrary point on curve \( C \) and above the X-axis and \( H \) be a point on \( MF_1 \) such that \( MF_2 \perp F_1 F_2, \, M F_1 \perp OH, \, |OH| = \lambda |O F_2| \) with \( \lambda \in (2/5, 3/5) \), then the range of the eccentricity \( e \) is in:
Let the line $\frac{x}{4} + \frac{y}{2} = 1$ meet the x-axis and y-axis at A and B, respectively. M is the midpoint of side AB, and M' is the image of the point M across the line $x + y = 1$. Let the point P lie on the line $x + y = 1$ such that $\Delta ABP$ is an isosceles triangle with $AP = BP$. Then the distance between M' and P is: