Find the vector equation of the plane with intercepts 3, –4 and 2 on x, y and z-axis respectively.
It is given that,
Intercepts are a = 3, b = −4, c = 2
The intercept form of a plane is as follows:
The equation of the given plane is
4x − 3y + 6z = 12.
`vec r.(4 hati-3hatj+6hatk)=12`
This is the vector form of the equation of the given plane.