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(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.

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