A variable plane frac{x}{a}+frac{y}{b}+frac{z | vedclass

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(A) The equation of the plane is $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$. Since it is at a unit distance from the origin $(0, 0, 0)$,we have $\frac{|-

Vedclass · Accepted Answer

$9$

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(A) The equation of the plane is $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$. Since it is at a unit distance from the origin $(0, 0, 0)$,we have $\frac{|-1|}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}}=1$,which implies $\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=1$. The coordinates of the vertices of $	riangle ABC$ are $A(a, 0, 0)$,$B(0, b, 0)$,and $C(0, 0, c)$. The centroid $(x, y, z)$ of $	riangle ABC$ is given by $\left(\frac{a}{3}, \frac{b}{3}, \frac{c}{3}ight)$. Thus,$x = \frac{a}{3}$,$y = \frac{b}{3}$,and $z = \frac{c}{3}$,which means $a = 3x$,$b = 3y$,and $c = 3z$. Substituting these into the condition $\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=1$,we get $\frac{1}{(3x)^2}+\frac{1}{(3y)^2}+\frac{1}{(3z)^2}=1$. This simplifies to $\frac{1}{9x^2}+\frac{1}{9y^2}+\frac{1}{9z^2}=1$,or $\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=9$. Comparing this with $\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=k$,we find $k=9$.

$A$ variable plane $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$,which is at a unit distance from the origin,cuts the coordinate axes at $A, B$,and $C$. If the centroid $(x, y, z)$ of $\triangle ABC$ satisfies $\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=k$,then $k$ equals:

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