Question #da508

1 Answer
Jun 11, 2017

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 ) #