Question

Find the vector equation of the plane with intercepts 3, –4 and 2 on x, y and z-axis respectively.

Answer 1

It is given that,
Intercepts are a = 3, b = −4, c = 2
The intercept form of a plane is as follows:

`x/a+y/b+z/c=1`

`therefore x/3+y/-4+z/2=1`

The equation of the given plane is
4x − 3y + 6z = 12.

`(xhati+yhatj+zhatk).(4hati-3hatj+6hatk)=12`

`vec r.(4 hati-3hatj+6hatk)=12`

This is the vector form of the equation of the given plane.