A point moves in such a way that the sum of i | vedclass

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

Question

(C) Let the coordinates of the point be $P(x, y, z)$.The distance of point $P(x, y, z)$ from the $xy$-plane is $|z|$.The distance of point $P(x, y, z)

Vedclass · Accepted Answer

$x - y + z = 0$

Vedclass · Answer

(C) Let the coordinates of the point be $P(x, y, z)$.The distance of point $P(x, y, z)$ from the $xy$-plane is $|z|$.The distance of point $P(x, y, z)$ from the $yz$-plane is $|x|$.The distance of point $P(x, y, z)$ from the $zx$-plane is $|y|$.According to the problem,the sum of its distances from the $xy$-plane and $yz$-plane is equal to its distance from the $zx$-plane:$|z| + |x| = |y|$.Assuming the point lies in the first octant where $x, y, z > 0$,we have $z + x = y$.Rearranging the terms,we get $x - y + z = 0$.

$A$ point moves in such a way that the sum of its distances from the $xy$-plane and $yz$-plane remains equal to its distance from the $zx$-plane. The locus of the point is:

Similar Questions

If $A(3, 5), B(-5, -4), C(7, 10)$ are the vertices of a parallelogram,taken in order,then the coordinates of the fourth vertex are:

The mid-points of the sides of a triangle are $(5,7,11)$,$(0,8,5)$,and $(2,3,-1)$. Find its vertices.

If $D(2, 1, 0)$, $E(2, 0, 0)$, and $F(0, 1, 0)$ are the mid-points of the sides $BC$, $CA$, and $AB$ of $\triangle ABC$, respectively, then the centroid of $\triangle ABC$ is:

If the midpoints of the sides $AB$, $BC$, and $CA$ of a triangle are respectively $D(1, 2, -3)$, $E(3, 0, 1)$, and $F(-1, 1, -4)$, then the centroid of the triangle $ADF$ is

If $D(2, 1, 0)$,$E(2, 0, 0)$,and $F(0, 1, 0)$ are mid-points of the sides $BC$,$CA$,and $AB$ of $\triangle ABC$,respectively,then the centroid of $\triangle ABC$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module