Find the parametric equations for the line of intersection of the planes x+y+z=3 and x-y+2z=2
I took the cross product of the 2 equations and got 3i-j-2k
I then set z=0 and got x=5/2 and y=1/2.
I got:
x=5/2 +3t
y=1/2-t
z=-2t
However, the answers are supposed to be:
x=5/2-(3/2)t
y=1/2+(1/2)t
z=t
What is my procedure missing to get there?
I think your method is correct.
You took the cross product of the two vectors of the plane <1,1,1>x<1,-1,2> and got the direction vector v=<3,-1,-2>
You combined the two equations and got (0,5/2,1/2), which gives you the position
There are many different solutions for an equation of a line. So don't think that your answer is wrong.
*oops (5/2,1/2,0) for position
Looking at the answer again,
it looks like they divided the vector by 2
Do you know why they divided it by 2?
To find the parametric equations for the line of intersection of the planes x+y+z=3 and x-y+2z=2, there are a few steps missing in your procedure. Let's go through the correct steps:
1. Start with the two given equations for the planes:
- Plane 1: x + y + z = 3
- Plane 2: x - y + 2z = 2
2. To find the line of intersection, we need to solve the system of equations formed by the two planes. We can use the method of substitution or elimination to do this.
3. Let's use the elimination method to eliminate x. To do so, subtract the second equation from the first equation:
- (x + y + z) - (x - y + 2z) = 3 - 2
- x - x + y - (-y) + z - 2z = 1
- 2y - z = 1
4. Now, we have two equations:
- y + z = 3 (Equation 1)
- 2y - z = 1 (Equation 2)
5. Solve the system of equations (Equation 1 and Equation 2) to find the values of y and z. To do so, you can either use substitution or elimination. For simplicity, let's use the elimination method:
- Multiply Equation 2 by 2:
4y - 2z = 2
- Add Equation 1 and the modified Equation 2:
(y + z) + (4y - 2z) = 3 + 2
5y - z = 5
- Now, you have a new equation:
5y - z = 5 (Equation 3)
6. We now have two equations:
- 2y - z = 1 (Equation 2)
- 5y - z = 5 (Equation 3)
7. Solve the new system of equations using the elimination method:
- Multiply Equation 2 by 5 and Equation 3 by 2:
10y - 5z = 5
10y - 2z = 10
- Subtract the modified Equation 2 from the modified Equation 3:
(10y - 2z) - (10y - 5z) = 10 - 5
3z = 5
- Solve for z:
z = 5/3
8. Substitute the value of z back into either Equation 2 or Equation 3 to solve for y. Let's use Equation 2:
2y - (5/3) = 1
2y = 1 + 5/3
2y = 8/3
y = (8/3)(1/2)
y = 4/3
9. We now have the values of y and z. Substitute these values into one of the original equations (let's use Equation 1) to solve for x:
y + z = 3
(4/3) + (5/3) = 3
x = 3 - (4/3) - (5/3)
x = (9/3) - (4/3) - (5/3)
x = 0
10. The point of intersection of the two planes is (0, 4/3, 5/3).
11. To find the parametric equations for the line of intersection, let t be a parameter:
x = 0 + 0t = 0
y = (4/3) + (1/3)t
z = (5/3) + (1/3)t
Therefore, the correct parametric equations for the line of intersection are:
x = 0
y = (4/3) + (1/3)t
z = (5/3) + (1/3)t
Note: The equation you obtained, 3i - j - 2k, is the direction vector of the line of intersection, but it is not the parametric equation itself.