Question

Find the equation of the plane whose intercepts on the axes \(x, y, z\) are respectively \(2, 3\), and \(-4\).

Hint:

Use the intercept form of the plane equation: \(\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1\).

Answer 1

The equation of a plane with intercepts \(a, b, c\) on the \(x, y, z\) axes respectively is: \[ \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1. \] Substitute \(a=2, b=3, c=-4\): \[ \frac{x}{2} + \frac{y}{3} + \frac{z}{-4} = 1. \] Multiply both sides by 12 to clear denominators: \[ 6x + 4y - 3z = 12. \]
Final answer: \[ \boxed{ 6x + 4y - 3z = 12. } \]