(Vectors question) section im on is Relationships between points, lines and planes

[Namrak]

Delevop a cartesian equation of a plane with x-intercept a, y-intercept b and z-intercept c.
Show that the distance d from the origin to this plane is given by 1/d^2=1/a^2+1/b^2+1/c^2.

Work ive done
d=√(a^2+b^2+c^2)
n->(normal)=(a,b,c)
p->(point P)=(0,0,0)
a->(any point on plane)=(a,0,0)
Pa->=(a,0,0)

 

[pka]

Delevop a cartesian equation of a plane with x-intercept a, y-intercept b and z-intercept c.
Show that the distance d from the origin to this plane is given by 1/d^2=1/a^2+1/b^2+1/c^2.

The normal to the plane is \(\displaystyle N=<-a,b,0>\times<-a,0,c>\). Let \(\displaystyle \vec{A}=<a,0,0>\).

Then the distance you want is \(\displaystyle \dfrac{|\vec{A}\cdot N|}{\|N\|}\).