difficult analytic geometry question because it has x y z.?

Question

Can you explain to me how to answer this question. Please do it clearly.

1. Find the equation of a sphere having
(5, -3, -2) & (0, 4 , -1) as ends of a diameter.

2. A moving point is always equidistant from the point
(3, 6 , -1) and the plane x = -3. Find the equation its locus.

3. The distance of a moving point from the plane Z = 4 is always half its distance from the point (1, -4, 5). Find the equation of the locus

Answer 1

#1: If the two given points are the endpoints of a diameter, then the center of the sphere must be their midpoint, which is (5/2, 1/2, -3/2). The radius of the sphere is the distance from the midpoint to either of the two given points, as both of them lie on the sphere itself. Thus r = √((5/2)² + (7/2)² + (1/2)²) = 1/2 √(25 + 49 + 1) = 1/2 √75 = 5√3/2. Plugging this into the equation of a sphere yields (x-5/2)² + (y-1/2)² + (z+3/2)² = 75/4.

#2: We know that the point's distance from the plane x=-3 is simply |x+3| and the distance formula yields that the distance from (3, 6, -1) is √((x-3)² + (y-6)² + (z+1)²). Setting these equal to each other and squaring both sides yields that:

(x+3)² = (x-3)² + (y-6)² + (z+1)²
x²+6x+9 = x²-6x+9 + (y-6)² + (z+1)²
12x = (y-6)² + (z+1)²
x = ((y-6)² + (z+1)²)/12

You can expand that out further if you want, but you can already see that it is the equation of an elliptic paraboloid.

#3: Same techniques as the previous problem, except this time you will end up with the equation of a hyperboloid.

Answer 2

( 1 ) The equation of the sphere with (x', y', z') and (x", y", z") as ends of a diameter is given as

(x - x')(x - x") + (y - y')(y - y") + ( z - z')(z - z") = 0.

Here, (x', y', z') = (5, -3, -2) and (x", y", z") = (0, 4, - 1 )

Hence, the equation of the sphere is

(x - 5)(x - 0) + (y -+3)(y - 4) + (z + 2)(z + 1) = 0, or

x^2 + y^2 + z^2 - 5x - y + 3z - 10 = 0.

( 2 ) Let the moving point be ( x, y, z )

Distances are equal means square of the distances are equal.

Square of its distance from the given point (3, 6, - 1) is

(x - 3)^2 + (y - 6)^2 + (z + 1 )^2 ... ... ( 1 )

Square of its distance from the plane x = - 3 is

(x + 3)^2 ... ... ( 2 )

Equatting ( 1 ) and ( 2 ) and simplifying we get the equation of the locus as

y^2 + z^2 - 12x - 12y + 2z + 37 = 0

( 3 ) Try this yourself taking hint from the previous problem.

Answer 3

1) centre of the sphere is; (2.5, 0.5, -1.5)
radious = root79/2
eq, (x-2.5)^2 +(y-0.5)^2 +(z +1.5)^2= 79/2

2)let the pt is (l,m,n)
then according to the condition,
(l-3)^2+(m-6)^2+(n+1)^2= (l+3)^2+m^2 +n^2
then the locus will be just replace (l,m,n) by ( x,y,z) after simplifying the eq

3) let the pt (l,m,n)
then according to the condition,
l^2 + m^2 + (n-4)^2= [(l-1)^2 + (m+4)^2+ ( n-5)^2]/2
simplify the eq and replace (l,m,n) by (x,y,z) as abve

Answer 4

dia = sqr [ 5^2 + 7^2 + 1^2 ] = 5 sqr(3)
center point
x = 5/2
y = 1/2
z = -3/2
equation of sphere
(x-5/2)^2 + (y-1/2)^2 + (z+3/2)^2 = [ 5/2sqr(3) ] ^2 = 75/4