(x-1)^2+(y-2)^2+(z-3)^2=9 represents a sphere in 3Dspace. Find the equation in form Ax+By+Cz+D=0..see detail.?

Question

The equation (x-1)^2+(y-2)^2+(z-3)^2=9 represents a sphere in 3D space. Find the equation in the form of Ax+By+Cz+D=0 of the plance that is tangent to the sphere at (2,4,5), a point at one end of the diameter of the sphere.

Can someone please explain to me in detail, i've not learn how to solve this in class.

Thanks.

Oh and also,
How do you find the equation of the sphere whith centre (2,1,3) and tangent to the plane 3x-4y+2z=12.

Im really confused that this point, but i managed to start the question and am stuck now.

Answer 1

OK, think about a sphere in 3D and a plane tangent to it at a point. The line from the centre of the sphere to the tangent point is normal to the plane - that's the key to this stuff. Also, remember that a plane in form Ax + By + Cz + D = 0 has normal vector (A, B, C).

For the first question, the centre of the sphere is at (1, 2, 3) and the tangent point is (2, 4, 5), so the vector from one to the other is (1, 2, 2). This is normal to the tangent plane, so the tangent plane has form
x + 2y + 2z + D = 0
and contains the point (2, 4, 5), so put this in:
2 + 2(4) + 2(5) + D = 0 => D = -20. So the plane is
x + 2y + 2z - 20 = 0.

For the second one, all we need to find is the radius of the sphere. There is a long but straightforward way to do it, and a shorter but trickier way. Here's the first way:

Let the tangent point be (x, y, z). Then we know the vector (x-2, y-1, z-3) and the vector (3, -4, 2) are both normal to the plane, so they must be scalar multiples:
(x-2, y-1, z-3) = k(3, -4, 2) for some k
=> x = 3k + 2, y = -4k + 1, z = 2k + 3
We also know it is on the plane, so 3x - 4y + 2z = 12
=> 3(3k + 2) - 4(-4k + 1) + 2(2k + 3) = 12
=> 9k + 6 + 16k - 4 + 4k + 6 = 12
=> 29k = 4, k = 4/29
So x-2 = 12/29, y-1 = -16/29, z-3 = 8/29.

The equation of the sphere is
(x-2)^2 + (y-1)^2 + (z-3)^2 = r^2
Substituting in we get r^2 = (12/29)^2 + (-16/29)^2 + (8/29)^2
= 464 / 841 = 16/29.
So the full equation is
(x-2)^2 + (y-1)^2 + (z-3)^2 = 16/29.

For the shorter method, note that we only need to find the radius of the sphere to have the full equation; this will be the distance from the centre of the sphere to the plane.
Let n = (3, -4, 2), so the plane has vector equation n . x = 12. Now ||n|| = sqrt(3^2 + (-4)^2 + 2^2) = sqrt(29). Moving distance 1 along the direction of n will give an increase of sqrt(29) in the value of n . x. At the centre of the sphere n . x = 6 - 4 + 6 = 8, so we need to increase this by 4 to get to the plane. This will require moving a distance 4 / sqrt(29), which is therefore the radius of the sphere. So the equation is
(x-2)^2 + (y-1)^2 + (z-3)^2 = (4 / sqrt(29))^2 = 16/29.

Answer 2

3y + 2Z = 12 y - z = 9. from the 2d: y = z+9 exchange into the 1st 3(z+9) + 2z = 12 3z + 27 + 2z = 12 5z + 27 = 12 subtract 27 from the two facets 5z = -15 divide by using 5 z = -3 y = z+9 = -3+9 = 6 the respond is C.

Answer 3

Dude.

Put the pencil down and walk away from the desk.

What could this possibly solve?

I dont get it-adding nubers with letters-WHAT- just add numbers.

why ad them anyway.

Take a Break

Answer 4

Talk to your teacher

Answer 5

This is for...? It would make more sense if I knew where this was all coming from.