If the sum of the squares of the distances of | vedclass

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

Question

(B) The distance of point $P(x, y, z)$ from the $x$-axis is $\sqrt{y^2 + z^2}$.The distance of point $P(x, y, z)$ from the $y$-axis is $\sqrt{x^2 + z^

Vedclass · Accepted Answer

$11$

Vedclass · Answer

(B) The distance of point $P(x, y, z)$ from the $x$-axis is $\sqrt{y^2 + z^2}$.The distance of point $P(x, y, z)$ from the $y$-axis is $\sqrt{x^2 + z^2}$.The distance of point $P(x, y, z)$ from the $z$-axis is $\sqrt{x^2 + y^2}$.According to the problem,the sum of the squares of these distances is $242$:$(y^2 + z^2) + (x^2 + z^2) + (x^2 + y^2) = 242$$2(x^2 + y^2 + z^2) = 242$$x^2 + y^2 + z^2 = 121$The distance of point $P(x, y, z)$ from the origin $(0, 0, 0)$ is given by $d = \sqrt{x^2 + y^2 + z^2}$.Substituting the value,$d = \sqrt{121} = 11$.Thus,the distance is $11$ units.

If the sum of the squares of the distances of the point $P(x, y, z)$ from the coordinate axes is $242$,then the distance of the point $P$ from the origin is . . . . . . units.

Similar Questions

How far apart are the points $(2,0,0)$ and $(-3,0,0)$?

Find the distance between the following pairs of points: $(-3, 7, 2)$ and $(2, 4, -1)$.

The coordinates of a point which is equidistant from the points $(0, 0, 0), (a, 0, 0), (0, b, 0)$ and $(0, 0, c)$ are given by

The distance of the point $P(a, b, c)$ from the $x$-axis is

If the extremities of a diagonal of a square are $(1, -2, 3)$ and $(2, -3, 5)$,then the length of its side is

Vedclass Test Series

Exam Paper Generator

Online Exam Module