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Question

The radius of circular section in which the sphere |r|=5 is cut by the plane r(^i+^j+^k)=33, is equal to

A
4.00
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B
4
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C
4.0
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Solution

Let A be the foot of perpendicular from the centre O to the plane r(^i+^j+^k)33=0

Then, |OA|=∣ ∣0(^i+^j+^k)33|^i+^j+^k|∣ ∣=333=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, OP= radius of sphere =5
Now, AP2=OP2OA2=5232=16
AP=4

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