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!
Sure, I can help you break down and solve the problem step by step.
(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 plane $ is <1, 2, -2>. Using this normal vector, we can form the equation of the plane @ by substituting the point (1, 1, 2) into the equation of the plane and solve for the constant term.
So, let's substitute the values into the equation for $:
x + 2y - 2z - 6 = 1 + 2 - 4 - 6 = -7
The equation for plane @ is: x + 2y - 2z - 7 = 0.
(a)(ii) To calculate the distance between planes $ and @, we can use the formula for the distance between a point and a plane. Let's use the formula:
Distance = | ax + by + cz + d | / sqrt(a^2 + b^2 + c^2)
In this case, the point (1, 1, 2) lies on the plane @ (since we constructed @ to contain this point), so we substitute the values into the equation for plane $:
Distance = |1(1) + 2(1) - 2(2) - 6| / sqrt(1^2 + 2^2 + (-2)^2)
= |-5| / sqrt(9)
= 5 / 3
Therefore, the distance between plane $ and plane @ is 5/3 units.
(b)(i) To find the parametric equations for the line of intersection of planes $ and %, we need to find the direction vector of the line. The direction vector is the cross product of the normal vectors of the two planes.
The normal vector of plane $ is <1, 2, -2>.
The normal vector of plane % is <2, -1, 1>.
Taking the cross product of these two normal vectors, we get the direction vector of the line of intersection:
Direction vector = <1, 2, -2> x <2, -1, 1> =
= <3, 4, 5>
Now, we can use the parametric equation of a line to write the equations:
x = x_0 + at
y = y_0 + bt
z = z_0 + ct
where (x_0, y_0, z_0) is a point on the line of intersection and (a, b, c) is the direction vector.
We can choose any point on the line. Let's choose a convenient point, for example, the point of intersection of planes $ and %. To find this point, we need to solve the simultaneous equations:
x+ 2y -2z -6 = 0 (equation for plane $)
2x - y + z + 8 = 0 (equation for plane %)
Solving these equations simultaneously, we find:
x = -2, y = 4, z = 0
Therefore, the parametric equations for the line of intersection are:
x = -2 + 3t
y = 4 + 4t
z = 5t
(b)(ii) To find the equations of the spheres tangent to £, radius 2, and whose centers lie on the line with the equation x = -2 + 0€, y = 4 + 1€, z = 0 + 1€, we need to find the center coordinates of the spheres.
From the given equation for the line, we can obtain the center coordinates of the spheres using the parameter "€" as follows:
Center coordinates = (-2, 4 + 1€, 0 + 1€)
Since the radius of the sphere is given as 2, the general equation of a sphere is:
(x - a)^2 + (y - b)^2 + (z - c)^2 = r^2
where (a, b, c) are the center coordinates and r is the radius.
Substituting the center coordinates and radius into the equation, we get:
(x + 2)^2 + (y - (4 + 1€))^2 + (z - (0 + 1€))^2 = 2^2
(x + 2)^2 + (y - 4 - €)^2 + (z - €)^2 = 4
Therefore, the equation of the spheres tangent to £, with radius 2 and centers lying on the line x = -2 + 0€, y = 4 + 1€, z = 0 + 1€, is given by:
(x + 2)^2 + (y - 4 - €)^2 + (z - €)^2 = 4