Vector/Geometry toughie 2

1. Mar 29, 2008

emma3001

If a, b, and c are the x, y and z intercepts of a plane respectively and d is the distance from the origin to the plane, prove that:

1/dsquared = 1/a squared + 1/b squared + 1/c squared

A (a, 0, 0)
B (0, b, 0)
C (0, 0, c)

Then I made vector AB [-a, b, 0] and vector BC [0, -b, c]. Then I found the cross product of AB x BC, which gave me a normal vector of [bc, ac, ab]. If these are my A, B and C for the scalar equation of a plane, then:

bcx + acy + abz + D= 0
I solved for D, which is -abc.

Therefore, the scalar equation for the plane is

bcx + acy + abz -abc= 0

Now I am completely stuck as how to use that scalar equation to prove the above equation.

2. Mar 29, 2008

tiny-tim

Hi emma!

Hint: [bc,ca,ab] = abc[1/a,1/b,1/c].

3. Apr 1, 2008

emma3001

thank you tiny-tim...