A plane meets the coordinate axes $A, B, C$ $A, B, C$ such that the centroid of the triangle $A B C$ $ABC$ is the point $(a, b, c)$ $(a, b, c)$ , show that the equation of the plane is $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 3$ $x/a+y/b+z/c=3$ ?

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Answer 1 · 2017-07-07T09:24:20

See the proof below

The equation of the plane is

$\frac{x}{A} + \frac{y}{B} + \frac{z}{C} = 1$ $x/A+y/B+z/C=1$

By the definition of the centroid

$(\frac{A}{3}, \frac{B}{3}, \frac{C}{3}) = (a, b, c)$ $(A/3,B/3,C/3)=(a,b,c)$

Therefore,

$A = 3 a$ $A=3a$

$B = 3 b$ $B=3b$

$C = 3 c$ $C=3c$

The equation of the plane becomes

$\frac{x}{3 a} + \frac{y}{3 b} + \frac{z}{3 c} = 1$ $x/(3a)+y/(3b)+z/(3c)=1$

$\frac{1}{3} (\frac{x}{a} + \frac{y}{b} + \frac{z}{c}) = 1$ $1/3(x/a+y/b+z/c)=1$

So,

$\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 3$ $x/a+y/b+z/c=3$

A plane meets the coordinate axes A, B, CA,B,CA, B, C such that the centroid of the triangle ABCABCABC is the point (a, b, c)(a,b,c)(a, b, c), show that the equation of the plane is x/a+y/b+z/c=3xa+yb+zc=3x/a+y/b+z/c=3?