# Question da508

Jun 11, 2017

Intersection coordinate is $\left(2 , 1 , 0\right)$

#### Explanation:

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

${\Pi}_{1} : x + 2 y - z = 4$
${\Pi}_{2} : 3 x - y + z = 5$
${\Pi}_{3} : 2 x + 3 y + 2 z = 7$

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:

$- 24 z = 0 \implies z = 0$

From Row $2$ we have:

$- 7 y + 4 z = - 7 \implies - 7 y = - 7 \implies y = 1$

From Row $1$ we have:

$x + 2 y - z = 4 \implies x + 2 - 0 = 4 \implies x = 2$

Thus we have a unique solution:

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

Making the coordinate of intersection of the planes:

$\left(2 , 1 , 0\right)$