A plane meets the coordinate axes at points A | vedclass

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(A) Let the coordinates of points $A$, $B$, and $C$ be $(a, 0, 0)$, $(0, b, 0)$, and $(0, 0, c)$ respectively.The centroid of triangle $ABC$ is $(\fra

Vedclass · Accepted Answer

$\frac{6}{\sqrt{26}}$

Vedclass · Answer

(A) Let the coordinates of points $A$, $B$, and $C$ be $(a, 0, 0)$, $(0, b, 0)$, and $(0, 0, c)$ respectively.The centroid of triangle $ABC$ is $(\frac{a}{3}, \frac{b}{3}, \frac{c}{3})$.Given the centroid is $(2, -\frac{2}{3}, \frac{1}{2})$, we have $\frac{a}{3} = 2 \implies a = 6$, $\frac{b}{3} = -\frac{2}{3} \implies b = -2$, and $\frac{c}{3} = \frac{1}{2} \implies c = \frac{3}{2}$.The equation of the plane is $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$, which becomes $\frac{x}{6} + \frac{y}{-2} + \frac{z}{3/2} = 1$, or $\frac{x}{6} - \frac{y}{2} + \frac{2z}{3} = 1$.Multiplying by $6$, we get $x - 3y + 4z - 6 = 0$.The perpendicular distance $d$ from the origin $(0, 0, 0)$ to the plane $Ax + By + Cz + D = 0$ is $d = \frac{|D|}{\sqrt{A^2 + B^2 + C^2}}$.Here, $d = \frac{|-6|}{\sqrt{1^2 + (-3)^2 + 4^2}} = \frac{6}{\sqrt{1 + 9 + 16}} = \frac{6}{\sqrt{26}}$.

$A$ plane meets the coordinate axes at points $A$, $B$ and $C$, such that the centroid of triangle $ABC$ is $(2, -\frac{2}{3}, \frac{1}{2})$. The perpendicular distance from the origin to this plane is:

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