A variable plane at a constant distance p fro | vedclass

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

Question

(A) Let the equation of the plane be $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$,where $a, b, c$ are the intercepts on the $x, y, z$ axes respective

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let the equation of the plane be $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$,where $a, b, c$ are the intercepts on the $x, y, z$ axes respectively.The distance of this plane from the origin $(0, 0, 0)$ is given by $p = \frac{|-1|}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}}}$.Squaring both sides,we get $p^2 = \frac{1}{\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}}$,which implies $\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} = \frac{1}{p^2}$.Since the planes are drawn parallel to the coordinate planes through $A(a, 0, 0)$,$B(0, b, 0)$,and $C(0, 0, c)$,the point of intersection of these planes is $(a, b, c)$.Replacing $(a, b, c)$ with $(x, y, z)$,the locus is $\frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2} = \frac{1}{p^2}$.

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

Similar Questions

In each of the following cases,determine the direction cosines of the normal to the plane and the distance from the origin: $5y + 8 = 0$.

The position vectors of two points $P$ and $Q$ are $3i + j + 2k$ and $i - 2j - 4k$ respectively. The equation of the plane passing through $Q$ and perpendicular to $PQ$ is:

Find the Cartesian equation of the following plane:
$\vec{r} \cdot [(s-2t) \hat{i} + (3-t) \hat{j} + (2s+t) \hat{k}] = 15$

The Cartesian equation of a plane parallel to the plane $\vec{r} \cdot(2 \hat{i}+3 \hat{j}-4 \hat{k})=1$ and at a distance of $2$ units from it is

If $a, b, c$ are the intercepts made on $X, Y, Z$-axes respectively by the plane passing through the points $(1, 0, -2), (3, -1, 2)$ and $(0, -3, 4)$, then $3a + 4b + 7c =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module