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
To find the parametric equations for the line of intersection of the planes x+y+z=3 and x-y+2z=2, you need to solve the system of equations simultaneously.
First, let's solve the system by using the method of elimination:
Given equations:
1) x + y + z = 3
2) x - y + 2z = 2
To eliminate x, let's subtract equation (2) from equation (1):
(x + y + z) - (x - y + 2z) = 3 - 2
x + y + z - x + y - 2z = 1
2y - z = 1
Now, we have two equations:
1) x + y + z = 3
2) 2y - z = 1
To eliminate z, let's multiply equation (1) by 2 and add it to equation (2):
2(x + y + z) + 2y - z = 6 + 1
2x + 3y = 7
Now, we have two equations:
1) x + y + z = 3
2) 2x + 3y = 7
To further eliminate x, let's multiply equation (1) by 2 and subtract it from equation (2):
2(2x + 3y) - (x + y + z) = 14 - 3
4x + 6y - x - y - z = 11
3x + 5y - z = 11
Now, we have two equations:
1) x + y + z = 3
2) 3x + 5y - z = 11
To eliminate z, let's add equation (1) to equation (2):
(x + y + z) + (3x + 5y - z) = 3 + 11
4x + 6y = 14
Now, we have one equation:
4x + 6y = 14
To find the parameterization, solve this equation for x and y:
4x + 6y = 14
2x + 3y = 7
Let's set y as the parameter t:
2x + 3t = 7
2x = 7 - 3t
x = 7/2 - (3/2)t
Now, we can substitute this value of x into the equation 1) x + y + z = 3:
(7/2 - (3/2)t) + y + z = 3
y + z = 3 - (7/2) + (3/2)t
y + z = 6/2 - 7/2 + (3/2)t
y + z = (-1/2) + (3/2)t
Let's substitute y + z as another parameter s:
s = (-1/2) + (3/2)t
Now, we can express y and z in terms of the parameter s:
y = s - z
y = [(-1/2) + (3/2)t] - z
y = (-1/2) + (3/2)t -z
Substitute z as another parameter w:
y = (-1/2) + (3/2)t - w
Finally, we can express the line of intersection as parametric equations:
x = 7/2 - (3/2)t
y = (-1/2) + (3/2)t - w
z = w
These are the correct parametric equations for the line of intersection of the given planes.
To find the parametric equations for the line of intersection of the planes x+y+z=3 and x-y+2z=2, you correctly found the direction vector of the line by taking the cross product of the normal vectors of the two planes.
The cross product of the normal vectors (coefficients of x, y, and z) of the two planes can be calculated as follows:
(i) Multiply the coefficients of y and z in the first plane equation by -1.
(ii) Multiply the coefficients of x and z in the first plane equation by -1.
(iii) Multiply the coefficients of x, y, and z in the second plane equation by -1 and 2 respectively.
(iv) Add the corresponding terms.
Following these steps, the cross product becomes 3i - j - 2k, which is the direction vector of the line.
Now we need to find a point on the line in order to write the parametric equations. You chose to set z=0, which is a valid choice. However, it seems like you made a mistake in the substitution.
Let's go through the steps again:
(i) Set z = 0. Then the equations become:
x + y = 3
x - y = 2
(ii) Solve the system of equations to find x and y:
Adding the two equations, we get 2x = 5, hence x = 5/2.
Substituting x back into either equation, we get y = 1/2.
So one point on the line is (5/2, 1/2, 0).
Finally, we can write the parametric equations for the line:
x = 5/2 + 3t
y = 1/2 - t
z = -2t
However, note that your answers x = 5/2 - (3/2)t, y = 1/2 + (1/2)t, and z = t are also correct. The parametric equations for a line can have different parameterizations as long as they represent the same line.