Question

Question #da508

Answer 1

Intersection coordinate is `( 2, 1, 0 )`

Explanation:

The planes will meet at a simultaneous solution to their equations:

`Pi_1: x+2y-z=4`
`Pi_2: 3x-y+z=5`
`Pi_3: 2x+3y+2z=7`

We can solve this system of linear equations by using Gaussian Elimination by setting up an augmented matrix of the equation coefficients.

`( (1, 2, -1, |, 4), (3, -1, 1, |, 5), (2, 3, 2, |, 7) )`

We can now perform elementary row operations:

`( (1, 2, -1, |, 4), (3, -1, 1, |, 5), (2, 3, 2, |, 7) ) stackrel(R_2-3R_1 rarr R_2)(rarr) ( (1, 2, -1, |, 4), (0, -7, 4, |, -7), (2, 3, 2, |, 7) )`

`( (1, 2, -1, |, 4), (0, -7, 4, |, -7), (2, 3, 2, |, 7) ) stackrel(R_3-2R_1 rarr R_3)(rarr) ( (1, 2, -1, |, 4), (0, -7, 4, |, -7), (0, -1, 4, |, -1) )`

`( (1, 2, -1, |, 4), (0, -7, 4, |, -7), (0, -1, 4, |, -1) ) stackrel(R_2-7R_3 rarr R_3)(rarr) ( (1, 2, -1, |, 4), (0, -7, 4, |, -7), (0, 0, -24, |, 0) )`

We can now use back substitution to get the values of `x`, `y`, and `z`:

From Row `3` we have:

`-24z = 0 => z = 0`

From Row `2` we have:

`-7y+4z=-7 => -7y=-7 => y=1`

From Row `1` we have:

`x + 2y-z = 4 => x +2-0 =4 => x=2`

Thus we have a unique solution:

`x=2, y=1, z=0`

Making the coordinate of intersection of the planes:

`( 2, 1, 0 )`