A plane cuts the coordinate axes X, Y, Z at A | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Let the equation of the plane be $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$. Since the plane cuts the axes at $A(a, 0, 0)$,$B(0, b, 0)$,and $C(

Vedclass · Accepted Answer

$x+y+2z-18=0$

Vedclass · Answer

(D) Let the equation of the plane be $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$. Since the plane cuts the axes at $A(a, 0, 0)$,$B(0, b, 0)$,and $C(0, 0, c)$,the centroid of $	riangle ABC$ is given by $\left(\frac{a+0+0}{3}, \frac{0+b+0}{3}, \frac{0+0+c}{3}ight) = \left(\frac{a}{3}, \frac{b}{3}, \frac{c}{3}ight)$. Given that the centroid is $(6, 6, 3)$,we have: $\frac{a}{3} = 6 \Rightarrow a = 18$ $\frac{b}{3} = 6 \Rightarrow b = 18$ $\frac{c}{3} = 3 \Rightarrow c = 9$ Substituting these values into the plane equation: $\frac{x}{18} + \frac{y}{18} + \frac{z}{9} = 1$ Multiplying by $18$,we get: $x + y + 2z = 18$ $x + y + 2z - 18 = 0$.

$A$ plane cuts the coordinate axes $X, Y, Z$ at $A, B, C$ respectively such that the centroid of the $\triangle ABC$ is $(6, 6, 3)$. Then the equation of that plane is

Similar Questions

Find the angle between the planes $ax + by + d = 0$ $(a^2 + b^2 \neq 0)$ and $z = 0$.

The equation of the plane passing through $(-1, 1, 2)$,whose normal makes equal acute angles with the coordinate axes is

$A$ variable plane at a constant distance $p$ from the origin meets the coordinate axes at $A, B, C$. Through these points,planes are drawn parallel to the coordinate planes. The locus of the point of intersection is

In three-dimensional space,what does the equation $3y + 4z = 0$ represent?

The distance between two parallel planes $2x + y + 2z = 8$ and $4x + 2y + 4z + 5 = 0$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module