Two systems of rectangular axes have the same | vedclass

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

Question

(D) Let the equation of the plane in the first system of rectangular axes be $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$.The perpendicular distance

Vedclass · Accepted Answer

$\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} - \frac{1}{a'^2} - \frac{1}{b'^2} - \frac{1}{c'^2} = 0$

Vedclass · Answer

(D) Let the equation of the plane in the first system of rectangular axes be $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$.The perpendicular distance $p$ from the origin $(0, 0, 0)$ to this plane is given by $p = \frac{|-1|}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}}}$,which implies $\frac{1}{p^2} = \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}$.Similarly,for the second system of rectangular axes,the equation of the same plane is $\frac{x}{a'} + \frac{y}{b'} + \frac{z}{c'} = 1$.The perpendicular distance $p$ from the origin to this plane is the same,so $p = \frac{|-1|}{\sqrt{\frac{1}{a'^2} + \frac{1}{b'^2} + \frac{1}{c'^2}}}$,which implies $\frac{1}{p^2} = \frac{1}{a'^2} + \frac{1}{b'^2} + \frac{1}{c'^2}$.Equating the two expressions for $\frac{1}{p^2}$,we get $\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} = \frac{1}{a'^2} + \frac{1}{b'^2} + \frac{1}{c'^2}$.Therefore,$\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} - \frac{1}{a'^2} - \frac{1}{b'^2} - \frac{1}{c'^2} = 0$.

Two systems of rectangular axes have the same origin. If a plane cuts them at distances $a, b, c$ and $a', b', c'$ from the origin along the axes,then:

Similar Questions

$A$ plane meets the coordinate axes at $P, Q,$ and $R$ such that the position vector of the centroid of $\Delta PQR$ is $2i - 5j + 8k$. Then the equation of the plane is:

If the foot of the perpendicular drawn from the origin to a plane is $M(-1, -2, 2)$,then the vector equation of the plane is

The plane passing through $(2, 1, -3)$ and perpendicular to $3 \hat{i} - \hat{j} + 2 \hat{k}$ contains the points

Distance between the two planes: $2x + 3y + 4z = 4$ and $4x + 6y + 8z = 12$ is

The plane which bisects the line segment joining the points $A(4, -2, 3)$ and $B(2, 4, -1)$ at right angles also passes through the point:

Vedclass Test Series

Exam Paper Generator

Online Exam Module