what is equation of sphere when diameter points are given ?
namku:
i used the bisector formula to find centre point of the line joining the two given points n used that as the centre of the circle but the signs are coming wrong
that should work, what exactly are the diameter end points ?
1,2,3 and 3,4,5
what do you get for center and radius ?
2,3,4 centre
looks good, what about radius
26
can u solve n tell by both methods ? m not getting my mistake
(2, 1, 4) and (4, 3, 10)
what ?
Answers
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center: (2,3,4)
diameter = sqrt[(3-1)^2 + (4-2)^2 + (5-3)^2]
diameter = sqrt(2^2 + 2^2 + 2^2) = sqrt(4 + 4 + 4) =sqrt(12)
radius = sqrt(12) / 2
we can leave it like this, since the RHS of the equation will be r^2, or 12/4
equation ==> (x - 2)^2 + (y - 3)^2 + (z - 4)^2 = 12/4
diameter = sqrt[(3-1)^2 + (4-2)^2 + (5-3)^2]
diameter = sqrt(2^2 + 2^2 + 2^2) = sqrt(4 + 4 + 4) =sqrt(12)
radius = sqrt(12) / 2
we can leave it like this, since the RHS of the equation will be r^2, or 12/4
equation ==> (x - 2)^2 + (y - 3)^2 + (z - 4)^2 = 12/4
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