A plane meets the co-ordinate axes in A, B, C such that the centroid of the triangle ABC is the point ( 1 , r , r 2 ) , then the equation of the plane is,

Question

TestHub · Accepted Answer

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 equation of the plane in intercept form is $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$ .

The centroid of the triangle $A BC$ is given by $(\frac{a}{3}, \frac{b}{3}, \frac{c}{3})$ .

Given that the centroid is $(1, r, r^{2})$ , we have:

$\frac{a}{3} = 1 \Rightarrow a = 3$

$\frac{b}{3} = r \Rightarrow b = 3 r$

$\frac{c}{3} = r^{2} \Rightarrow c = 3 r^{2}$

Substituting the values of $a$ , $b$ , and $c$ into the equation of the plane, we get:

$\frac{x}{3} + \frac{y}{3 r} + \frac{z}{3 r ^{2}} = 1$

Multiplying the entire equation by $3 r^{2}$ , we obtain:

$r^{2} x + ry + z = 3 r^{2}$

Answer: $r^{2} x + ry + z = 3 r^{2}$

Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

Options:

Doubts & Discussion