Question

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

Answer 1

Hint:The most important thing in this question is that we should know the formula of centroid of the triangle ABC. For x coordinate it is like $x = \dfrac{{{x_1} + {x_2} + {x_3}}}{3}$. In this question we just need to substitute the appropriate values of different variables in the formula.


Formula used:
Centroid of triangle ABC is $\left( {a,b,c} \right)$.
Therefore, $a = \dfrac{{{x_1} + {x_2} + {x_3}}}{3}$ ,$b = \dfrac{{{y_1} + {y_2} + {y_3}}}{3}$ , $c = \dfrac{{{z_1} + {z_2} + {z_3}}}{3}$

Complete step by step answer:
Centroid of the $\vartriangle ABC$ is $\left( {a,b,c} \right)$.
Equation of the plane in the intercept form is
$\dfrac{x}{p} + \dfrac{y}{q} + \dfrac{z}{r} = 1$ where p, q, r are x-intercept, y-intercept, z-intercept respectively.
Now, we will find the centroid of the triangle.
Using the formula $a = \dfrac{{{x_1} + {x_2} + {x_3}}}{3}$ ,$b = \dfrac{{{y_1} + {y_2} + {y_3}}}{3}$ , $c = \dfrac{{{z_1} + {z_2} + {z_3}}}{3}$
Now substitute the values as ${x_1} = p\,,\,{x_2} = 0\,,\,{x_3} = 0$, ${y_1} = 0\,,\,{y_2} = q,\,{y_3} = 0$, ${z_1} = 0\,,\,{z_2} = 0\,,\,{z_3} = r$
On substituting the values, we get,
$a = \dfrac{{p + 0 + 0}}{3} = \dfrac{p}{3}............\left( 1 \right)$
$b = \dfrac{{0 + q + 0}}{3} = \dfrac{q}{3}.............\left( 2 \right)$
$c = \dfrac{{0 + 0 + r}}{3} = \dfrac{r}{3}...............\left( 3 \right)$
$\dfrac{x}{p} + \dfrac{y}{q} + \dfrac{z}{r} = 1$
Now we can say that $p = 3a$, $q = 3b$ and $r = 3c$.
We know that equation of plane is $\dfrac{x}{p} + \dfrac{y}{q} + \dfrac{z}{r} = 1$
Now substitute the values of p, q, r in the above equation.
$\dfrac{x}{{3a}} + \dfrac{y}{{3b}} + \dfrac{z}{{3c}} = 1$
Now taking out the common $3$ from the denominator.
$\dfrac{1}{3}\left( {\dfrac{x}{a} + \dfrac{y}{b} + \dfrac{z}{c}} \right) = 1$
Now on cross-multiplication, we get
$\dfrac{x}{a} + \dfrac{y}{b} + \dfrac{z}{c} = 3$ is our required equation.
Therefore, the equation is $\dfrac{x}{a} + \dfrac{y}{b} + \dfrac{z}{c} = 3$.

Note: In the above question we have used the value of ${y_1}\,and\,{z_1}$ equals to zero because they are present on the x-axis. Similarly, we have use the value of ${x_2}\,and\,{z_2}$ equals to zero because they are present on y-axis and the value of ${x_3}\,and\,{y_3}$ equals to zero because they are present on z-axis.