Find the radius of the circular section of the sphere x 2 + y2 + z2 = 49 by the plane 2x + 3y – z – 5 14 = 0
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Answer:
Correct option is A)
Centre of sphere x
2
+y
2
+z
2
=49 is the origin (0,0,0) and radius r=7
Let p= length of ⊥ er from (0,0,0) to plane 2x+3y−z=5
14
=
14
∣2×03×0−0−5
15
∣
=5
∴R=
r
2
−p
2
=
7
2
−5
2
=2
6
sorry this is explanation
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