#### Question

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

#### Solution

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

4*x* − 3*y* + 6*z* = 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.

Is there an error in this question or solution?

Solution Find the Vector Equation of the Plane with Intercepts 3, –4 and 2 on X, Y and Z-axis Respectively. Concept: Vector and Cartesian Equation of a Plane.