What is the nature of intersection of these three planes:
5x+3y-z=11
-2x-7y+10z=2
6x+21y-30z=-6
I know that the answer is two parallel planes and the other one intersecting in a line. my question is how would you know that the first equation intersects in a line? Would you use elementary operations?
correction, the two planes are coincident that the other intersects in line
unless the normals are parallel, the planes must intersect in a line.
To determine the nature of the intersection of the three planes, you can use elementary operations to put the system of equations in a simplified form called the "Row Echelon Form" or "Reduced Row Echelon Form."
Here's how you can proceed:
1. Write the system of equations in the form AX = B:
5x + 3y - z = 11 (equation 1)
-2x - 7y + 10z = 2 (equation 2)
6x + 21y - 30z = -6 (equation 3)
2. Create the augmented matrix [A|B] using the coefficients of the variables:
[5 3 -1 | 11]
[-2 -7 10 | 2]
[6 21 -30 | -6]
3. Apply elementary row operations to transform the augmented matrix into the row echelon form. These operations include:
- Swapping two rows
- Multiplying a row by a non-zero constant
- Adding or subtracting a multiple of one row from another
Continue with the following steps until the matrix is in row echelon form:
4. Make the leading coefficient (the leftmost non-zero entry) in the first row (row 1) equal to 1 by multiplying row 1 by 1/5:
[1 3/5 -1/5 | 11/5]
[-2 -7 10 | 2]
[6 21 -30 | -6]
5. Use row 1 to eliminate the leading coefficient in row 2 by multiplying row 1 by 2 and adding to row 2:
[1 3/5 -1/5 | 11/5]
[0 -49/5 8/5 | 24/5]
[6 21 -30 | -6]
6. Use row 1 to eliminate the leading coefficient in row 3 by multiplying row 1 by -6 and adding to row 3:
[1 3/5 -1/5 | 11/5]
[0 -49/5 8/5 | 24/5]
[0 11/5 -23/5 | -44/5]
7. Make the leading coefficient in row 2 equal to 1 by multiplying row 2 by -(5/49):
[1 3/5 -1/5 |11/5]
[0 1 -8/49 | -24/49]
[0 11/5 -23/5 | -44/5]
8. Use row 2 to eliminate the leading coefficient in row 3 by multiplying row 2 by -11/5 and adding to row 3:
[1 3/5 -1/5 |11/5]
[0 1 -8/49 | -24/49]
[0 0 1 | 0]
9. Now, the augmented matrix is in the row echelon form. The equation corresponding to the last row is "0x + 0y + 1z = 0," simplifying to "z = 0."
So, the system of equations simplifies to:
1. x + (3/5)y - (1/5)z = 11/5
2. y - (8/49)z = -24/49
3. z = 0
This means we have one equation in terms of x and y and one equation which only involves z. The variable z is independent in this case and can have any real value. Performing back-substitution, we can solve for x and y in terms of z:
From equation 3, we have z = 0.
Substituting z = 0 into equation 2, we get y - (8/49)(0) = -24/49, which simplifies to y = -24/49.
Substituting z = 0 and y = -24/49 into equation 1, we get x + (3/5)(-24/49) - (1/5)(0) = 11/5, which simplifies to x = 11/5 - (3/5)(-24/49).
Therefore, the intersection of the three planes is a line, where x = 11/5 - (3/5)(-24/49), y = -24/49, and z = 0.