You are given the following two planes:
x+4y-3z-12=0
x+6y-2z-22=0
a) Determine if the planes are parallel
b) Fine the line of intersection of the two planes.
c) Use the two original equations to determine two other equations that have the same solution as the original two.
d) Find a third equation that will have a unique solution with the original two equations.
To determine whether the given planes are parallel, we need to compare their normal vectors. The normal vector of a plane can be found by extracting the coefficients of the variables x, y, and z in the equation of the plane.
Let's find the normal vector for each plane:
For the first plane, x + 4y - 3z - 12 = 0, the coefficients of x, y, and z are 1, 4, and -3, respectively. So the normal vector of the first plane is (1, 4, -3).
For the second plane, x + 6y - 2z - 22 = 0, the coefficients of x, y, and z are 1, 6, and -2, respectively. So the normal vector of the second plane is (1, 6, -2).
a) Since the normal vectors of the two planes are not scalar multiples of each other, i.e., they are not proportional, we can conclude that the planes are not parallel.
b) To find the line of intersection of the two planes, we need to find the direction vector of the line. This can be done by taking the cross product of the normal vectors of the planes.
Calculating the cross product:
(1, 4, -3) × (1, 6, -2) = ((4 * -2) - (6 * -3), (-3 * 1) - (-2 * 1), (1 * 6) - (4 * 1))
= (-8 + 18, -3 + 2, 6 - 4)
= (10, -1, 2)
So the direction vector of the line of intersection is (10, -1, 2).
Now, we need a point on the line. This can be obtained by simultaneously solving the two plane equations. Let's use an elimination method:
Multiply the first equation by 2: 2(x + 4y - 3z - 12) = 0
Multiply the second equation by 3: 3(x + 6y - 2z - 22) = 0
Simplifying the equations:
2x + 8y - 6z - 24 = 0
3x + 18y - 6z - 66 = 0
Subtracting the first equation from the second equation:
(3x - 2x) + (18y - 8y) + (-6z + 6z) - (66 + 24) = 0
x + 10y - 90 = 0
Since x can be any value, let's set x = 0:
0 + 10y - 90 = 0
10y = 90
y = 9
Substituting x = 0 and y = 9 into the first plane equation:
0 + 4(9) - 3z - 12 = 0
36 - 3z - 12 = 0
-3z = -24
z = 8
So a point on the line of intersection is (0, 9, 8).
Therefore, the line of intersection of the two planes is given by the equation:
L: (x, y, z) = (0, 9, 8) + t(10, -1, 2), where t is a parameter.
c) To determine two other equations that have the same solution as the original two, we can multiply both planes by a non-zero scalar value. Let's choose 2 as the scalar:
2(x + 4y - 3z - 12) = 0
2(x + 6y - 2z - 22) = 0
Simplifying the equations:
2x + 8y - 6z - 24 = 0
2x + 12y - 4z - 44 = 0
So the two other equations that have the same solution as the original ones are:
2x + 8y - 6z - 24 = 0
2x + 12y - 4z - 44 = 0
d) To find a third equation that will have a unique solution with the original two equations, we can use any equation that is not a scalar multiple of the other two. Let's create a new equation by adding the equations of the two given planes:
(x + 4y - 3z - 12) + (x + 6y - 2z - 22) = 0
2x + 10y - 5z - 34 = 0
So the third equation that will have a unique solution with the original two equations is:
2x + 10y - 5z - 34 = 0.