A variable plane is at a constant distance h | 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 $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$. The points where it meets the coordinate axes are $A(a, 0, 0)$,$B(

Vedclass · Accepted Answer

$\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=\frac{9}{h^2}$

Vedclass · Answer

(D) Let the equation of the plane be $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$. The points where it meets the coordinate axes are $A(a, 0, 0)$,$B(0, b, 0)$,and $C(0, 0, c)$.The distance of this plane from the origin $(0, 0, 0)$ is given as $h$. The formula for the distance is $\frac{1}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}}} = h$,which implies $\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} = \frac{1}{h^2}$.Let $(x, y, z)$ be the coordinates of the centroid of $	riangle ABC$. Then $x = \frac{a}{3}$,$y = \frac{b}{3}$,and $z = \frac{c}{3}$.This gives $a = 3x$,$b = 3y$,and $c = 3z$.Substituting these into the distance equation:$\frac{1}{(3x)^2} + \frac{1}{(3y)^2} + \frac{1}{(3z)^2} = \frac{1}{h^2}$$\frac{1}{9x^2} + \frac{1}{9y^2} + \frac{1}{9z^2} = \frac{1}{h^2}$$\frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2} = \frac{9}{h^2}$

$A$ 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

Similar Questions

Consider the following three planes:
$P : x + y - 2z + 7 = 0$
$Q : x + y + 2z + 2 = 0$
$R : 3x + 3y - 6z - 11 = 0$

Statement $-1:$ The point $A(3,1,6)$ is the mirror image of the point $B(1,3,4)$ in the plane $x-y+z=5.$
Statement $-2:$ The plane $x-y+z=5$ bisects the line segment joining $A(3,1,6)$ and $B(1,3,4).$

The perpendicular distance from the origin to the plane containing the points having position vectors $\hat{i}+2\hat{j}+3\hat{k}$,$2\hat{i}+3\hat{j}-4\hat{k}$,and $3\hat{i}-4\hat{j}+5\hat{k}$ is

The equation of the plane passing through $3 \hat{i}+2 \hat{j}+6 \hat{k}$ and parallel to the vectors $2 \hat{i}+\hat{j}+\hat{k}$ and $\hat{i}-\hat{j}+\hat{k}$ is

The equation of the plane which is parallel to the $xy$-plane and cuts an intercept of length $3$ from the $z$-axis is

Vedclass Test Series

Exam Paper Generator

Online Exam Module