Three planes:
x+y-2z=1
2x+z=y-4
-3x-y+z=-2
They intersect at: (a) (-2/5,5,9/5) (b) no common intersection (c) r=(5/3,-2/3,0)+&(1,5,3) (d) r=(0,-4,-5)+%(1,5,3)
&/%=signs like alpha and beta, used in vector eqns
Please help! I just can't get the answer to be/ressemble any of them!
To find the point of intersection between three planes, you need to solve the system of equations formed by the equations of the planes. Let's solve them step by step:
1. Write the system of equations in matrix form:
| 1 1 -2 | | 1 |
| 2 -1 1 | | -4 |
| -3 -1 1 | = | -2 |
2. Use Gaussian elimination or any other preferred method to transform the augmented matrix into row-echelon form:
| 1 1 -2 | | 1 |
| 0 -3 5 | | -6 |
| 0 0 0 | = | 1 |
3. The third row of the row-echelon form shows that 0x + 0y + 0z = 1, which is not possible. This means that the system of equations is inconsistent, and there is no common intersection point for the three planes. Therefore, the correct answer is (b) no common intersection.
It seems that none of the given options (a), (c), or (d) match the solution because there is no common intersection for the given system of equations.