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
a. If the planes are parallel, one equation would be a linear combination of the other, which is not the case.
b. as in the previous problem,
http://www.jiskha.com/display.cgi?id=1283324887
reduce the equations to the reduced echelon form, namely
1 0 -5 = -8
0 2 1 = 10
Set z=t (free variable) and solve for x, y and z to get
x=5t-8
y=-(1/2)t+5
z=t
To determine if the given planes are parallel, we need to compare their normal vectors. The normal vector of a plane is the coefficients of the variables (x, y, and z) in its equation.
For the first plane, x + 4y - 3z - 12 = 0, the normal vector is (1, 4, -3).
For the second plane, x + 6y - 2z - 22 = 0, the normal vector is (1, 6, -2).
Now, if the two planes are parallel, their normal vectors should be scalar multiples of each other. In other words, one normal vector should be obtainable by multiplying the other vector by a constant.
To check if the normal vectors are scalar multiples, we can find the ratio of their corresponding components:
(1/1) = 4/6 = -3/-2
Since the ratios are equal, we can conclude that the normal vectors are scalar multiples of each other, which means that the planes are parallel.
Now, to find the line of intersection of the two planes, we can solve the system of equations formed by setting the equations of the planes equal to each other:
x + 4y - 3z - 12 = x + 6y - 2z - 22
Rearranging the equation, we get:
2y - z = 10
To express this equation parametrically, we can assign a variable t to either y or z and express the other variable in terms of t. Let's choose t = y.
Using t = y, we can express y as t and substitute it back into the original equation:
2t - z = 10
Rearranging this equation to solve for z, we get:
z = 2t - 10
Now that we have z in terms of t, we can substitute it back into one of the original equations to find x in terms of t. Let's use the first equation:
x + 4y - 3z - 12 = 0
Substituting y = t and z = 2t - 10, we get:
x + 4t - 3(2t - 10) - 12 = 0
Simplifying this equation, we get:
x - 2t + 30 = 0
Rearranging this equation to solve for x, we get:
x = 2t - 30
So, the line of intersection of the two planes is given by the parametric equations:
x = 2t - 30
y = t
z = 2t - 10