You are given the following two planes:
x+4y-3z-12=0
x+6y-2z-22=0
a) Use the two original equations to determine two other equations that have the same solution as the original two.
ANS: the orginal two are not parallel in anyway so can I just take multiples of the original two OR is there any other way to find them?
b) Find a third equation that will have a unique solution with the original two equations.
ANS: 3x + 2y - z - 18 = 0
a) One way to determine two other equations that have the same solution as the original two is by taking multiples of the original equations. Let's multiply the first equation by 2 and the second equation by 3:
2(x + 4y - 3z - 12) = 0
3(x + 6y - 2z - 22) = 0
Simplifying these equations gives us:
2x + 8y - 6z - 24 = 0
3x + 18y - 6z - 66 = 0
These two equations have the same solution as the original two.
b) To find a third equation that will have a unique solution with the original two equations, we can use the method of elimination. Let's eliminate one variable from the two equations by multiplying the first equation by 3 and the second equation by 4:
3(x + 4y - 3z - 12) = 0
4(x + 6y - 2z - 22) = 0
Simplifying these equations gives us:
3x + 12y - 9z - 36 = 0
4x + 24y - 8z - 88 = 0
Now we can subtract the first equation from the second equation:
(4x + 24y - 8z - 88) - (3x + 12y - 9z - 36) = 0
Simplifying this equation gives us:
x + 12y + z - 52 = 0
Therefore, the third equation that will have a unique solution with the original two equations is x + 12y + z - 52 = 0.
a) To determine two other equations that have the same solution as the original two, we can use the concept of linear combination.
1. Multiply the first equation by a constant, let's say 2, and the second equation by a different constant, let's say 3.
Equation 1: 2(x + 4y - 3z - 12) = 0 ==> 2x + 8y - 6z - 24 = 0
Equation 2: 3(x + 6y - 2z - 22) = 0 ==> 3x + 18y - 6z - 66 = 0
Now we have two additional equations that have the same solution as the original two.
b) To find a third equation that will have a unique solution with the original two equations, we need to find the intersection (common solution) of the given planes.
1. Solve the system of the original two equations to find the point of intersection.
Subtract equation 1 from equation 2:
(x + 6y - 2z - 22) - (x + 4y - 3z - 12) = 0
Simplifying, we get:
2y + z = -4
Now, substitute this value back into either of the original equations (let's choose equation 1):
x + 4y - 3z - 12 = 0
x + 4y - 3(-4 - 2y) - 12 = 0
x + 4y + 6y + 12 - 12 = 0
x + 10y = 0
So, the third equation that will have a unique solution with the original two equations is:
2y + z = -4
x + 10y = 0
Therefore, the third equation is 2y + z = -4 and x + 10y = 0.