- #1

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**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/

**d**squared = 1/

**a**squared + 1/

**b**squared + 1/

**c**squared

I made three points

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.