Question 36362: (a): Find the equation of the line joining the points
(─5, 2, 3) and (5, ─ 2, 3)
(b) Find the equation of the sphere, which contains the circle
x^2 + y^2 + z^2 = 9, 3x + 3y + 3z = 5 and
passes through the origin.
Answer by khwang(438)   (Show Source):
You can put this solution on YOUR website! Don't poste duplicatively the same quetion.
(a): Find the equation of the line joining the points
(─5, 2, 3) and (5, ─ 2, 3)
Use parameter equation for a line directly
(x,y,z) = (-5,2,3) + ((5, ─ 2, 3) - (-5,2,3)) t
= (-5,2,3) + (10, -4,0) t.
so x = -5 + 10t, y = 2 -4t, z= 3
(b) Find the equation of the sphere, which contains the circle
x^2 + y^2 + z^2 = 9, 3x + 3y + 3z = 5 and
passes through the origin.
Find the equation of the sphere, which contains the circle
x^2 + y^2 + z^2 = 9 …(1),
3x + 3y + 3z = 5…(2) and
passes through the origin
Assume the equation of the sphere as:(center (a,b,c), radius r)
(x-a)^+ (y-b)^2 + (z-c)^2 = r^2….(3)
The intersection of two spheres(1) & (3) should satisfy (2):
(1) – (3) = 2(ax+by+cz) = 9 – r^2 is the same as eq. (2):
So, 2a/3 = 2b/3=2c/3 = (9 – r^2)/5 = k, a=b=c = 3k/2.
We also know that (3) passing through (0,0,0), thus a^2+b^2 + c^2 = r^2.
To solve for k, we have , 3(3k/2) ^2 = 9 – 5k. Hence, 27k^2 + 20 k - 36 = 0.
k = [-10 +/- 4sqrt(67)]/27. And then , a = b=c= 3k/2, r^2 = 9-5k.
The required equation has two solutions as:
(x- 3k/2)^2 +(y- 3k/2)^2+ (z- 3k/2)^2 = 9-5k
.... for you to simplify.
Kenny
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