A plane passing through a point (2,2,2) cuts | 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 passes through $(2, 2, 2)$,we have $\frac{2}{a} + \

Vedclass · Accepted Answer

$\alpha + \beta + \gamma \ge \frac{9}{2}$

Vedclass · Answer

(A) Let the equation of the plane be $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$.Since the plane passes through $(2, 2, 2)$,we have $\frac{2}{a} + \frac{2}{b} + \frac{2}{c} = 1$,which implies $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{1}{2}$.The coordinates of $A, B, C$ are $(a, 0, 0), (0, b, 0), (0, 0, c)$ respectively.The centroid $P(\alpha, \beta, \gamma)$ of tetrahedron $OABC$ is given by $\alpha = \frac{0+a+0+0}{4} = \frac{a}{4}$,$\beta = \frac{b}{4}$,and $\gamma = \frac{c}{4}$.Thus,$a = 4\alpha, b = 4\beta, c = 4\gamma$.Substituting these into the plane equation: $\frac{1}{4\alpha} + \frac{1}{4\beta} + \frac{1}{4\gamma} = \frac{1}{2} \Rightarrow \frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} = 2$.Using the $A.M. \ge H.M.$ inequality for positive numbers $\alpha, \beta, \gamma$:$\frac{\alpha + \beta + \gamma}{3} \ge \frac{3}{\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma}} = \frac{3}{2}$.Therefore,$\alpha + \beta + \gamma \ge \frac{9}{2}$.

$A$ plane passing through a point $(2,2,2)$ cuts the positive semi-axes at $A$,$B$,and $C$. If $P(\alpha, \beta, \gamma)$ is the centroid of the tetrahedron $OABC$ (where $O$ is the origin),then select the correct option.

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