A variable plane is at a constant distance p | vedclass

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Question

(A) Let the equation of the plane be $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$. Since the plane is at a constant distance $p$ from the origin $(0,

Vedclass · Accepted Answer

$x^{-2} + y^{-2} + z^{-2} = 16p^{-2}$

Vedclass · Answer

(A) Let the equation of the plane be $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$. Since the plane is at a constant distance $p$ from the origin $(0, 0, 0)$,we have $p = \frac{|-1|}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}}}$. Squaring both sides,we get $\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} = \frac{1}{p^2}$ ..... $(i)$. The coordinates of $A, B, C$ are $(a, 0, 0), (0, b, 0)$ and $(0, 0, c)$ respectively. The centroid $(x, y, z)$ of the tetrahedron $OABC$ is given by $x = \frac{0+a+0+0}{4} = \frac{a}{4}$,$y = \frac{0+0+b+0}{4} = \frac{b}{4}$,and $z = \frac{0+0+0+c}{4} = \frac{c}{4}$. Thus,$a = 4x, b = 4y, c = 4z$. Substituting these into equation $(i)$,we get $\frac{1}{(4x)^2} + \frac{1}{(4y)^2} + \frac{1}{(4z)^2} = \frac{1}{p^2}$. $\frac{1}{16x^2} + \frac{1}{16y^2} + \frac{1}{16z^2} = \frac{1}{p^2}$. Multiplying by $16$,we get $x^{-2} + y^{-2} + z^{-2} = 16p^{-2}$.

$A$ variable plane is at a constant distance $p$ from the origin and meets the axes in $A, B$ and $C$. The locus of the centroid of the tetrahedron $OABC$ is

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