Prove that if a plane has the intercepts a, b, c and is at a distance p units from the origin, then $1 over straight a squared plus 1 over straight b squared plus 1 over straight c squared space equals space 1 over straight p squared.$

Name: Wired Faculty - The Best Learning App
Rating: 4.6 (1864 reviews)

Solution

The equation of plane with intercepts a, b, c is

straight x over straight a plus straight y over straight b plus straight z over straight c equals space 1 space space space space space space space space or space space space straight x over straight a plus straight y over straight b plus straight z over straight c minus 1 space equals space 0

From the given condition,

fraction numerator open vertical bar 0 plus 0 plus 0 minus 1 close vertical bar over denominator square root of begin display style 1 over straight a squared end style plus begin display style 1 over straight b squared end style plus begin display style 1 over straight c squared end style end root end fraction space equals space straight p

rightwards double arrow space space space space space space space space space space space square root of 1 over straight a squared plus 1 over straight b squared plus 1 over straight c squared end root space equals 1 over straight p rightwards double arrow space space space space space space space space space space 1 over straight a squared plus 1 over straight b squared plus 1 over straight c squared space equals space 1 over straight p squared

which is required condition.

For Daily Free Study Material Join wiredfaculty