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(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

$3\sqrt{2}$

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(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:

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