A plane meets the coordinate axes at A, B, C | vedclass

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

Question

(A) 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 centroid of triangle $ABC$ is g

Vedclass · Accepted Answer

$\frac{x}{\alpha} + \frac{y}{\beta} + \frac{z}{\gamma} = 3$

Vedclass · Answer

(A) 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 centroid of triangle $ABC$ is given by $(\frac{a+0+0}{3}, \frac{0+b+0}{3}, \frac{0+0+c}{3}) = (\frac{a}{3}, \frac{b}{3}, \frac{c}{3})$.Given that the centroid is $(\alpha, \beta, \gamma)$,we have $\frac{a}{3} = \alpha$,$\frac{b}{3} = \beta$,and $\frac{c}{3} = \gamma$.Thus,$a = 3\alpha$,$b = 3\beta$,and $c = 3\gamma$.The intercept form of the equation of the plane is $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$.Substituting the values of $a, b, c$,we get $\frac{x}{3\alpha} + \frac{y}{3\beta} + \frac{z}{3\gamma} = 1$.Multiplying both sides by $3$,we obtain $\frac{x}{\alpha} + \frac{y}{\beta} + \frac{z}{\gamma} = 3$.

$A$ plane meets the coordinate axes at $A, B, C$ and $(\alpha, \beta, \gamma)$ is the centroid of the triangle $ABC$. Then the equation of the plane is

Similar Questions

The planes: $2x - y + 4z = 5$ and $5x - 2.5y + 10z = 6$ are

$A$ variable plane is at a constant distance $h$ from the origin and meets the coordinate axes in $A, B, C$. The locus of the centroid of $\triangle ABC$ is

An equation of a plane parallel to the plane $x-2y+2z-5=0$ and which is at a distance of $1$ unit from the origin is:

The mirror image of the point $A(1, 2, 3)$ in a plane is $B\left(-\frac{7}{3}, -\frac{4}{3}, -\frac{1}{3}\right)$. Which of the following points lies on this plane?

If the plane passing through the points $(1, 2, 3)$,$(2, 3, 1)$,and $(3, 1, 2)$ is $a x + b y + c z = 1$,then $a + 2 b + 3 c = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module