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

 Let the equation of plane be Here a = 2,  b = 3, c = 4∴    equation of plane is or      6x + 4y + 3z = 12 Text SolutionSolution : It is given that the `x,y` and `z` intercepts of the plane are `2,3` and `4` respectively, i.e.,
`a=2`
`b=3`
`c=4`
Now, it is known that the equation of a plane with intercepts `a,b` and `c` is given by,
`x/a+y/b+z/c=1`
Substituting the values of `a,b` and `c` in the above,
`x/2+y/3+z/4=1`
`=>(6xxx+4xxy+3xxz)/12=1`
`=>(6x+4y+3z)/12=1`
`=>(6x+4y+3z)/12xx12=1xx12`
`=>6x+4y+3z=12`
Which is the required equation of plane. Last updated at Feb. 1, 2020 by  This video is only available for Teachoo black users Solve all your doubts with Teachoo Black (new monthly pack available now!) Example 19 Find the equation of the plane with intercepts 2, 3 and 4 on the x, y and z-axis respectively. The equation of a plane with intercepts 𝑎, b, c on x, y, and z – axis respectively is 𝒙/𝒂 + 𝒚/𝒃 + 𝒛/𝒄 = 1 Given, Intercept on x − axis = 2 ∴ 𝑎 = 2 Intercept on y − axis = 3 ∴ b = 3 Intercept on z – axis = 4 ∴ c = 4 Equation of plane is 𝒙/𝟐 + 𝒚/𝟑 + 𝒛/𝟒 = 1 6𝑥/12 + 4𝑦/12 + 3𝑧/12 = 1 (6𝑥 + 4𝑦 + 3𝑧)/12 = 1 6x + 4y + 3z = 12 