3 planes; $: x+2y-2z-6=0

%: 2x-y+z+8=0
£: 2x-y+2z+3=0

(a)(i)Find the cartesian equation for the plane @ parallel to $ and containing the point (1,1,2)
(ii)Calculate the distance between $ and @

(b)(i)Find the parametric equations for the line of intersection of the planes $ and %
(ii)Find the equations of the spheres tangent to £, radius 2 and whose centres lie on the line with the equation:
x=-2+0€
y=4+1€
z=0+1€

Really confused, so please help!

(a)(i) To find the equation of a plane parallel to $ and containing the point (1,1,2), we can use the fact that parallel planes have the same normal vector.

The normal vector of the plane $ is <1, 2, -2>.

Since the planes $ and @ are parallel, @ will have the same normal vector.

The equation of the plane @ is given by:
x + 2y - 2z + d = 0, where d is a constant.

To find the value of d, substitute the coordinates of the point (1,1,2):
1 + 2(1) - 2(2) + d = 0
1 + 2 - 4 + d = 0
d = 1

Therefore, the equation of the plane @ parallel to $ and containing the point (1,1,2) is:
x + 2y - 2z + 1 = 0

(ii) To calculate the distance between planes $ and @, we can use the formula for the distance between a plane and a point.

The distance between two parallel planes can be found by taking the absolute difference of the constant terms in the plane equations and dividing it by the square root of the sum of the squares of the coefficients of x, y, and z in one of the plane equations.

For planes $ and @, the constant terms in the plane equations are -6 and 1, respectively.

The coefficients of x, y, and z in the plane $ equation are 1, 2, and -2, respectively.

Using the formula, the distance between $ and @ is given by:
Distance = |1 - (-6)| / sqrt(1^2 + 2^2 + (-2)^2)
= 7 / sqrt(1 + 4 + 4)
= 7 / sqrt(9)
= 7 / 3
= 2.33 (approx)

Therefore, the distance between $ and @ is approximately 2.33 units.

(b)(i) To find the parametric equations for the line of intersection of the planes $ and %, we need to solve the system of equations formed by setting the two equations equal to each other:

x + 2y - 2z - 6 = 0 (Plane $)
2x - y + z + 8 = 0 (Plane %)

From the second equation, we can isolate z and express it in terms of x and y:

z = -2x + y - 8

Let x = t.

Substituting x = t in z = -2x + y - 8, we get:

z = -2t + y - 8

Therefore, the parametric equations for the line of intersection of planes $ and % are:

x = t
y = 2t + 8
z = -2t + y - 8 = -3t

(b)(ii) The equation of the line where the centers of the tangent spheres lie is given as:

x = -2 + 0ε
y = 4 + 1ε
z = 0 + 1ε

To find the equation of a sphere tangent to £, with radius 2 and whose center lies on this line, we can substitute the equations of the line into the general equation of a sphere:

(x - a)^2 + (y - b)^2 + (z - c)^2 = r^2

Substituting the equations of the line, we get:

((-2 + 0ε) - a)^2 + ((4 + 1ε) - b)^2 + ((0 + 1ε) - c)^2 = 2^2

Expanding and simplifying, we have:

(-2a + 2ε)^2 + (4 - b + ε)^2 + (1ε - c)^2 = 4

Simplifying further, we get the equations:

4a^2 + 4b^2 + c^2 - 4ε(2a - b + c) - 6ε^2 - 4a + 2b + 2c + 15 = 0

Therefore, the equations of the spheres tangent to £, with radius 2 and centers lying on the given line, are:

4a^2 + 4b^2 + c^2 - 4ε(2a - b + c) - 6ε^2 - 4a + 2b + 2c + 15 = 0

(a)(i) To find the cartesian equation for the plane @ parallel to $ and containing the point (1,1,2), we can use the following steps:

1. First, find the direction vector of the plane $ by considering the coefficients of x, y, and z in its equation. The direction vector will be [-1, 2, -2].
2. The vector [-1, 2, -2] is parallel to the plane $, so any vector parallel to $ can be obtained by multiplying this vector by a non-zero scalar. Let's call this scalar "k".
3. Now, consider the point (1,1,2) that lies on the plane @. Subtracting the coordinates of this point from the coordinates of any point on plane $ will give us the vector parallel to both planes. Let's call this vector "v".
v = [x - 1, y - 1, z - 2] = [k(-1), k(2), k(-2)].
4. Equate the components of the vectors v and [-1, 2, -2] to find the value of k.
k(-1) = -1, k(2) = 2, k(-2) = -2.
From the first equation, k = 1. Substitute this value of k into the other two equations to verify that they are satisfied.
5. Now, we have the vector v = [-1, 2, -2].
6. Finally, to obtain the cartesian equation of plane @, substitute the coordinates of the point (1,1,2) into the equation of a plane and use the direction vector v.
x + 2y - 2z + d = 0 (where d is a constant to be determined)
1 + 2(1) - 2(2) + d = 0
1 + 2 - 4 + d = 0
d = 1
Therefore, the cartesian equation for plane @ is: x + 2y - 2z + 1 = 0.

(a)(ii) To calculate the distance between planes $ and @, we can use the formula for the distance between a point and a plane. We will substitute a point from one plane into the equation of the other plane.
1. Choose a point on plane $, such as (1,0,3), and substitute its coordinates into the equation of plane @.
x + 2y - 2z + 1 = 0
1 + 2(0) - 2(3) + 1 = 0
-5 ≠ 0
2. Since -5 ≠ 0, the point (1,0,3) is not on the plane @.
3. Therefore, the distance between planes $ and @ is the distance between the point (1,0,3) and the plane @. Calculate the distance using the formula for the distance between a point and a plane.
Distance = |(1 + 2(0) - 2(3) + 1)/√(1^2 + 2^2 + (-2)^2)|
= |-3/3|
= 1.

(b)(i) To find the parametric equations for the line of intersection of planes $ and %, we can simultaneously solve their equations to get the values of x, y, and z.
1. Eliminate one variable by adding or subtracting the two equations. Let's eliminate x by subtracting the equation of % from the equation of $.
(x + 2y - 2z -6) - (2x - y + z + 8) = 0,
-x + 3y - 3z - 14 = 0.
2. Now, we have:
-x + 3y - 3z - 14 = 0 (equation 1: obtained from $ and %)
2x - y + z + 8 = 0 (equation 2: obtained from %)
3. Solve the system of equations 1 and 2 for y and z.
4. We can solve this system of equations by substitution or elimination.
From equation 1: -x = 14 - 3y + 3z,
x = -14 + 3y - 3z.
5. Substitute x into equation 2:
2(-14 + 3y - 3z) - y + z + 8 = 0,
-28 + 6y - 6z - y + z + 8 = 0,
5y - 5z - 20 = 0,
5(y - z - 4) = 0.
6. Solve the equation 5(y - z - 4) = 0 for y - z - 4 = 0.
y - z - 4 = 0
y = z + 4.
7. Now, we have:
x = -14 + 3y - 3z,
y = z + 4.
8. Substitute y = z + 4 into x = -14 + 3y - 3z to find x in terms of z.
x = -14 + 3(z + 4) - 3z,
x = -14 + 3z + 12 - 3z,
x = -2.
9. Therefore, the parametric equations for the line of intersection of planes $ and % are:
x = -2,
y = z + 4.

(b)(ii) To find the equations of the spheres tangent to £, we need to consider two cases:
Case 1: The line of the centers, given as x = -2 + 0€, y = 4 + 1€, z = 0 + 1€, intersects the plane £.
In this case, the sphere will be externally tangent to £.
Case 2: The line of the centers is parallel to the plane £.
In this case, the sphere will be internally tangent to £.

Let's consider each case separately:

Case 1:
1. Substitute the given values into the equation of the plane £ to determine if the line of centers intersects the plane.
2(-2 + 0€) - (4 + 1€) + 2(0 + 1€) + 3 = 0,
-4 + 2€ - 4 - € + 2€ + 3 = 0,
0 = 0.
2. Since the equation is true, the line of centers intersects the plane £ in this case.
3. Now, let's find the radius of the spheres. Given that the radius is 2, and the centers are given by x = -2 + 0€, y = 4 + 1€, and z = 0 + 1€.
The distance between the center (-2, 4, 0) and any point on the line of centers (-2, 4, 0) + t(0, 1, 1) is given by the formula:
Distance = √((x2 - x1)^2 + (y2 - y1)^2 + (z2 - z1)^2),
= √((-2 - -2)^2 + (4 - 4)^2 + (0 - 0)^2),
= 0.
4. Since the distance is 0, the radius of the sphere is equal to the distance between the center and the intersection point.
5. Therefore, the equation of the sphere is:
(x + 2)^2 + (y - 4)^2 + z^2 = 0.

Case 2:
1. In this case, the line of centers is parallel to the plane £. Therefore, no intersection occurs.
2. Since there is no intersection, there is no sphere tangent to the plane £ in this case.

Therefore, the equations of the spheres tangent to £, with a radius of 2 units and the centers lying on the line x = -2 + 0€, y = 4 + 1€, z = 0 + 1€, are given by:
Case 1: (x + 2)^2 + (y - 4)^2 + z^2 = 0.
Case 2: No sphere in this case.