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Three Dimensional Geometry

Question
CBSEENMA12033484

A variable plane is at a constant distance p from the origin and meets the axes in A, B and C respectively, then show that locus of the centroid of the triangle ABC is
1 over straight x squared plus 1 over straight y squared plus 1 over straight z squared space equals space 9 over straight p squared.

Solution

Let O be the origin and OA = a, OB.= b, OC = c.
∴  the equation of plane passing through A, B and C is
                      straight x over straight a plus straight y over straight b plus straight z over straight c equals 1
or              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 space 1 over straight p
rightwards double arrow 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 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis
Now A, B, C are (a, 0,0), (0, b, 0), (0, 0, c) respectively.
Let (x1 ,y1 , z1) be the centroid of ΔABC.
 therefore space space space space space straight x subscript 1 space equals space fraction numerator straight a plus 0 plus 0 over denominator 3 end fraction comma space space space straight y subscript 1 space equals space fraction numerator 0 plus straight b plus 0 over denominator 3 end fraction comma space space straight z subscript 1 space equals space fraction numerator 0 plus 0 plus straight c over denominator 3 end fraction
therefore space space space space space space straight a space equals space 3 straight x subscript 1 comma space space space space straight b space equals space 3 space straight y subscript 1 comma space space space straight c space equals space 3 space straight z subscript 1
Putting values of a, b, c in (1), we get,
                      fraction numerator 1 over denominator 9 space straight x subscript 1 squared end fraction plus fraction numerator 1 over denominator 9 space straight y subscript 1 squared end fraction plus fraction numerator 1 over denominator 9 straight z subscript 1 squared end fraction space equals 1 over straight p squared
or                 1 over straight x subscript 1 squared plus 1 over straight y subscript 1 squared plus 1 over straight z subscript 1 squared space equals space 9 over straight p squared
∴  locus of centroid (x1 ,y1, z1) is
        1 over straight x squared plus 1 over straight y squared plus 1 over straight z squared space equals space 9 over straight p squared

Some More Questions From Three Dimensional Geometry Chapter

Find the direction cosines of x, y and z-axis.