Question

find the equation of planw which is parellel to plane 6x-3y-2z+9=0 and at distance 2 from origin

Answer 1

The normal-point form of  a plane passing through the point $(x_1,y_1,z_1)$ with a normal vector $n=<a,b,c>$   is ;

$a(x-x_1)+b(y-y_2)+c(z-z_1) = 0.$

From the plane  $6x-3y-2z=0$  we can read that the normal vector to the plane is $n=<6,-3,-2>$ . The plane parallel to this plane will have the same normal vector. We are free to choose the point that the plane goes through with the only restriction being that it should be 2 units away from the origin. For convenience,  I will pick the point (2,0,0). This point is 2 units away from the origin.

With this information, we have that ;

$a(x-x_1)+b(y-y_2)+c(z-z_1) = 0$

$6(x-2)-3(y-0)-2(z-0)=0\\6x-12-3y-2z=0\\6x-3y-2z-12=0$.

The plane $6x-3y-2z=12$ is paralle to the plane $6x-3y-2z+9=0$  and is two units away from the origin.

Answer 2

Answer:

Distance of a plane ax+by+cz+d=0 from origin is given by a2+b2+c2∣d∣

Here, distance from origin =62+32+2214

=4914

=714

=2