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) Let the point be $P(x, y, z)$.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 t

Vedclass · Accepted Answer

$3\sqrt{2}$

Vedclass · Answer

(B) Let the point be $P(x, y, z)$.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 $36$:$(\sqrt{y^2 + z^2})^2 + (\sqrt{x^2 + z^2})^2 + (\sqrt{x^2 + y^2})^2 = 36$$(y^2 + z^2) + (x^2 + z^2) + (x^2 + y^2) = 36$$2(x^2 + y^2 + z^2) = 36$$x^2 + y^2 + z^2 = 18$The distance of the point $P(x, y, z)$ from the origin $(0, 0, 0)$ is given by $\sqrt{x^2 + y^2 + z^2}$.Therefore,the distance is $\sqrt{18} = \sqrt{9 	imes 2} = 3\sqrt{2}$.

If the sum of the squares of the distances of a point from the three coordinate axes is $36$,then its distance from the origin is:

Similar Questions

If the coordinates of points $A$ and $B$ are $(3, 4, 5)$ and $(-1, 3, -7)$ respectively,find the equation of the locus of point $P(x, y, z)$ such that $PA^2 - PB^2 + 2k^2 = 0$.

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

Which of the following is at a distance of $\sqrt{10}$ from the point $(1, 2, 3)$?

The sum of the squares of the perpendicular distances of a point $(x, y, z)$ from the coordinate axes is $k$ times the square of the distance of the point from the origin. Then $k=$

The distance of the point $(1, 2, 3)$ from the $x$-axis is.......

Vedclass Test Series

Exam Paper Generator

Online Exam Module