The radius of circular section in which the sphere -r-5 is cut by the plane r-k-3-3- is equal to

Let A be the foot of perpendicular from the centre O to the plane nbsp;r·(î+ĵ+k̂) 3√(3)=0 Then, |OA|=|0·(î+ĵ+k̂) 3√(3)|î+ĵ+k̂||=3√(3)√(3)=3 (perpendicular di ...

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Standard XII

Mathematics

Equation of a Sphere : Diametric Form

The radius of...

Question

The radius of circular section in which the sphere $| \to r | = 5$ is cut by the plane $\to r \cdot (^i +^j +^k) = 3 \sqrt{3},$ is equal to

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Solution

Let $A$ be the foot of perpendicular from the centre $O$ to the plane $\to r \cdot (^i +^j +^k) - 3 \sqrt{3} = 0$

Then, $| O A | = ∣ ∣ ∣ ∣ \frac{0 \cdot (^i +^j +^k) - 3 \sqrt{3}}{|^i +^j +^k |} ∣ ∣ ∣ ∣ = \frac{3 \sqrt{3}}{\sqrt{3}} = 3$
(perpendicular distance of a point from the plane)
If, $P$ is any point on the circle, then $P$ lies on the plane as well as on the sphere,
So, $O P =$ radius of sphere $= 5$
Now, $A P^{2} = O P^{2} - O A^{2} = 5^{2} - 3^{2} = 16$
$\Rightarrow A P = 4$

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Equation of a Sphere : Diametric Form

Standard XII Mathematics

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The radius of circular section in which the sphere |→r|=5 is cut by the plane →r⋅(^i+^j+^k)=3√3, is equal to

The radius of circular section in which the sphere $| \to r | = 5$ is cut by the plane $\to r \cdot (^i +^j +^k) = 3 \sqrt{3},$ is equal to