Solve 2x-y+3z=8
x-6y-z=0
-6x+3y-9z=24
Did you notice that your third equation has the same normal as the third one?
So visually I see two parallel planes intersected by third non-parallel plane, namely the second equation.
So solve equations 1 and 2, and express them in parametric form
repeat by solving equation 2 and 3
these two lines will be parallel
hint: I got direction numbers of (19,5,-11) for the lines.
To solve the system of equations, we can first solve the equations 1 and 2 by eliminating either x, y, or z. In this case, we can choose to eliminate x.
From equation 2, we have: x - 6y - z = 0, which gives us x = 6y + z.
Substituting this into equation 1, we get:
2(6y + z) - y + 3z = 8
12y + 2z - y + 3z = 8
11y + 5z = 8
This is the first equation in terms of y and z.
Now, let's solve equations 2 and 3 to find the second equation in terms of y and z.
From equation 2: x - 6y - z = 0, we have x = 6y + z.
Substituting this into equation 3, we get:
-6(6y + z) + 3y - 9z = 24
-36y - 6z + 3y - 9z = 24
-33y - 15z = 24
This is the second equation in terms of y and z.
Now we have a system of two equations in the variables y and z:
11y + 5z = 8
-33y - 15z = 24
We can solve this system using any method, such as substitution or elimination. In this case, let's use elimination. Multiply the first equation by 3 and the second equation by -1 to make the coefficients of y the same:
33y + 15z = 24
33y + 15z = -24
Now, add the two equations together:
(33y + 15z) + (33y + 15z) = 24 + (-24)
66y + 30z = 0
Divide both sides by 6 to simplify the equation:
11y + 5z = 0
Now we have a new equation that represents the intersection between the two planes defined by equations 1 and 2.
To express this equation in parametric form, we can assign a parameter to either y or z. Let's assign a parameter t to z:
z = t
Substituting this into the equation 11y + 5z = 0, we get:
11y + 5(t) = 0
11y = -5t
y = (-5/11)t
So the parametric equation for the line formed by the intersection of the two planes defined by equations 1 and 2 is:
x = 6y + z = 6((-5/11)t) + t
x = (-30/11)t + t
x = (-30/11 + 11/11)t
x = (-19/11)t
To find the direction numbers of this line, we can compare the coefficients of t in the parametric equations. From the equation above, we can see that the direction numbers of the line are (-19/11, -5/11, 1).
Therefore, we have found the parametric equation and direction numbers of the line formed by the intersection of the two parallel planes defined by equations 1 and 2.