Question

A variable plane which remains at a constant distance 3p from the origin cuts the coordinate axes at a, b, c. Show that the locus of the centroid of triangle abc is

Answer 1

The locus of the centroid of Δabc is  $\frac{1}{p^{2} }$  $={(\frac{1}{x}) ^{2}+(\frac{1}{y} )^{2}+(\frac{1}{z} )^{2}}$.

To find : The locus of the centroid of triangle abc is $\frac{1}{x^{2} }+ \frac{1}{y^{2} } +\frac{1}{z^{2} } = \frac{1}{p^{2} }$.

Given :

Constant distance 3p from the origin cuts the co-ordinate axes at a, b, c.

Plane cuts the intercepts (a,b,c) on the axes is $\frac{x}{a}+ \frac{y}{b} +\frac{z}{c} = 1$.

With A (a, 0, 0), B (0, b, 0) and C (0, 0, c) distance of the plane from the origin is given by,

Centroid of ΔABC  $=\left(\frac{a+0+0}{3}, \frac{0+b+0}{3}, \frac{0+0+c}{3}\right)$        

                               $=\left(\frac{a}{3}, \frac{b}{3}, \frac{c}{3}\right)$

Distance between planes is given by,

                        D   $=\frac{\left|\frac{0}{a}+\frac{0}{b}+\frac{0}{c}-1\right|}{\sqrt{(\frac{1}{a}) ^{2}+(\frac{1}{b} )^{2}+(\frac{1}{c} )^{2}}}$                      

                              $=\frac{\left|-1|\right}{\sqrt{(\frac{1}{a}) ^{2}+(\frac{1}{b} )^{2}+(\frac{1}{c} )^{2}}}$

Length of the perpendicular from the origin (0, 0, 0) to the plane = 3 p.

                      3 p  $=\frac{\left1\right}{\sqrt{(\frac{1}{a}) ^{2}+(\frac{1}{b} )^{2}+(\frac{1}{c} )^{2}}}$                      

Squaring on both the sides, gives

                    $(3 p)^2$ $=(\frac{\left1\right}{\sqrt{(\frac{1}{a}) ^{2}+(\frac{1}{b} )^{2}+(\frac{1}{c} )^{2}}})^2$

While squaring, the square root and square get cancel each other.

                    9 p² $=\frac{\left1\right}{(\frac{1}{a}) ^{2}+(\frac{1}{b} )^{2}+(\frac{1}{c} )^{2}}$

                      $\frac{1}{9p^{2} }$  $={(\frac{1}{a}) ^{2}+(\frac{1}{b} )^{2}+(\frac{1}{c} )^{2}}$

Centroid of ΔABc is $\left(\frac{a}{3}, \frac{b}{3}, \frac{c}{3}\right)$ = ( x, y, z )

Here, a = 3x  ;  b = 3y  ;  c = 3z

                       $\frac{1}{9p^{2} }$  $={(\frac{1}{a}) ^{2}+(\frac{1}{b} )^{2}+(\frac{1}{c} )^{2}}$

                       $\frac{1}{9p^{2} }$  $={(\frac{1}{9x}) ^{2}+(\frac{1}{9y} )^{2}+(\frac{1}{9z} )^{2}}$

                        $\frac{1}{p^{2} }$  $={(\frac{1}{x}) ^{2}+(\frac{1}{y} )^{2}+(\frac{1}{z} )^{2}}$

Hence, the locus of the centroid of triangle abc is $\frac{1}{p^{2} }$  $={(\frac{1}{x}) ^{2}+(\frac{1}{y} )^{2}+(\frac{1}{z} )^{2}}$.

To learn more...

1. brainly.in/question/4081595                

2. brainly.in/question/1777678