A plane meets the coordinate axes at A, B, C | vedclass

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(A) Let the coordinates of the points where the plane meets the axes be $A(a, 0, 0)$,$B(0, b, 0)$,and $C(0, 0, c)$.The centroid of triangle $ABC$ is g

Vedclass · Accepted Answer

$\frac{x}{\alpha} + \frac{y}{\beta} + \frac{z}{\gamma} = 3$

Vedclass · Answer

(A) Let the coordinates of the points where the plane meets the axes be $A(a, 0, 0)$,$B(0, b, 0)$,and $C(0, 0, c)$.The centroid of triangle $ABC$ is given by $(\frac{a+0+0}{3}, \frac{0+b+0}{3}, \frac{0+0+c}{3}) = (\frac{a}{3}, \frac{b}{3}, \frac{c}{3})$.Given that the centroid is $(\alpha, \beta, \gamma)$,we have $\frac{a}{3} = \alpha$,$\frac{b}{3} = \beta$,and $\frac{c}{3} = \gamma$.Thus,$a = 3\alpha$,$b = 3\beta$,and $c = 3\gamma$.The intercept form of the equation of the plane is $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$.Substituting the values of $a, b, c$,we get $\frac{x}{3\alpha} + \frac{y}{3\beta} + \frac{z}{3\gamma} = 1$.Multiplying both sides by $3$,we obtain $\frac{x}{\alpha} + \frac{y}{\beta} + \frac{z}{\gamma} = 3$.

$A$ plane meets the coordinate axes at $A, B, C$ and $(\alpha, \beta, \gamma)$ is the centroid of the triangle $ABC$. Then the equation of the plane is

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