The radius of the circular section of the sph | vedclass

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(D) The equation of the sphere is $|r| = 5$,which represents a sphere with centre at the origin $(0, 0, 0)$ and radius $R = 5$.The given plane is $r \

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$4$

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(D) The equation of the sphere is $|r| = 5$,which represents a sphere with centre at the origin $(0, 0, 0)$ and radius $R = 5$.The given plane is $r \cdot (i + j + k) = 3\sqrt{3}$.Let $M$ be the foot of the perpendicular from the origin $O(0, 0, 0)$ to the plane. The length of the perpendicular $d$ from the origin to the plane $r \cdot n = p$ is given by $d = \frac{|p|}{|n|}$.Here,$p = 3\sqrt{3}$ and $n = i + j + k$,so $|n| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}$.Thus,$d = OM = \frac{3\sqrt{3}}{\sqrt{3}} = 3$.Let $r_{c}$ be the radius of the circular section. In the right-angled triangle $\Delta OPM$,where $P$ is a point on the circumference of the circular section,$OP$ is the radius of the sphere $(R = 5)$ and $OM$ is the distance from the centre to the plane $(d = 3)$.By the Pythagorean theorem,$OP^2 = OM^2 + r_{c}^2$.$5^2 = 3^2 + r_{c}^2$$25 = 9 + r_{c}^2$$r_{c}^2 = 16$$r_{c} = 4$.Therefore,the radius of the circular section is $4$.

The radius of the circular section of the sphere $|r| = 5$ by the plane $r \cdot (i + j + k) = 3\sqrt{3}$ is

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