If a,b and c are the x-intercept, the y-intercept and the z-intercept of a plane, respectively, and d is ...?

Question

If a,b and c are the x-intercept, the y-intercept and the z-intercept of a plane, respectively, and d is the distance from the origin to the plane, show that (1/d^2) = (1/ a^2) + (1/ b^2) + (1/c^2).

^2 -> square

please show all logical work and reasoning

Answer 1

The equation of a plane with those three intercepts is:

x/a + y/b + z/c = 1
x/a + y/b + z/c - 1 = 0

Let d = distance of plane from origin.

d = | (1/x)*0 + (1/b)*0 + (1/c)*0 - 1 | / √[(1/a)² + (1/b)² + (1/c)²]

d = 1 / √[(1/a)² + (1/b)² + (1/c)²]

d² = 1 / [(1/a)² + (1/b)² + (1/c)²]

1/d² = 1/a² + 1/b² + 1/c²

Answer 2

Your airplane will incorporate the standards (a,0,0), (0,b,0) and (0,0,c). as long as those are actually not co-linear, it incredibly is adequate preparation. we pick the path familiar to the airplane. If we variety a pair of vectors, we can take the bypass product to get the conventional. The airplane will incorporate the vectors with preliminary factor (a,0,0) and terminal factors (2 vectors right here) (0,b,0) and (0,0,c). those are u = <-a,b,0> v = <-a,0,c> Their bypass product is u x v = . The airplane containing (a,0,0) perpendicular to is bc(x-a) + acy + abz = 0 bcx + acy + abz = abc. This technique---the "non-colinearness"---calls for that a minimum of two of the values a, b, and c be nonzero.