Question

Find the equation of the plane with intercepts 2, 3 and 4 on the x, y and z-axes respectively

• A. 6x+4y+3z = 12
• B. 4x+3y-6z = 2
• C. x+y+z = 1
• D. 7x+2y+9z = 10

Answer 1

The correct option is A 6x+4y+3z = 12

The intercept form of a plane is given by $\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}+\frac{\text{z}}{\text{c}}=1$, a, b and c are the intercepts of the plane with x, y and z-axes respectively.
Here a = 2, b = 3, c = 4.
Substituting the values of a, b and c in (1), we get the required equation of the
plane as $\frac{\text{x}}{2}+\frac{\text{y}}{3}+\frac{\text{z}}{4}=1$ or 6x + 4y + 3z = 12.