We are asked to find the symmetric equations of the line of intersection of the two planes: $x + 4y - 2z = 13$ $2x - y - 2z = 5$

1. Problem Description

We are asked to find the symmetric equations of the line of intersection of the two planes:

$x + 4y - 2z = 13$

$2x - y - 2z = 5$

2. Solution Steps

First, we find the direction vector of the line of intersection. The direction vector is parallel to the cross product of the normal vectors of the two planes.

The normal vector of the first plane is $\vec{n_1} = <1, 4, -2>$.

The normal vector of the second plane is $\vec{n_2} = <2, -1, -2>$.

The direction vector $\vec{v}$ is given by:

$\vec{v} = \vec{n_1} \times \vec{n_2} = <(4)(-2) - (-1)(-2), (-2)(2) - (1)(-2), (1)(-1) - (4)(2)>$

$\vec{v} = <-8 - 2, -4 + 2, -1 - 8>$

$\vec{v} = <-10, -2, -9>$

Now, we need to find a point on the line of intersection. We can do this by setting one of the variables to a value (say, $z = 0$) and solving the resulting system of two equations in two variables.

$x + 4y = 13$

$2x - y = 5$

Multiply the second equation by 4:

$8x - 4y = 20$

Add this to the first equation:

$x + 4y + 8x - 4y = 13 + 20$

$9x = 33$

$x = \frac{33}{9} = \frac{11}{3}$

Substitute $x = \frac{11}{3}$ into $2x - y = 5$:

$2(\frac{11}{3}) - y = 5$

$\frac{22}{3} - y = 5$

$y = \frac{22}{3} - 5 = \frac{22 - 15}{3} = \frac{7}{3}$

So, a point on the line of intersection is $(\frac{11}{3}, \frac{7}{3}, 0)$.

The symmetric equations of the line are given by:

$\frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c}$

where $(x_0, y_0, z_0)$ is a point on the line and $<a, b, c>$ is the direction vector.

In this case, $(x_0, y_0, z_0) = (\frac{11}{3}, \frac{7}{3}, 0)$ and $<a, b, c> = <-10, -2, -9>$.

Thus, the symmetric equations are:

$\frac{x - \frac{11}{3}}{-10} = \frac{y - \frac{7}{3}}{-2} = \frac{z - 0}{-9}$

We can also use the direction vector $\langle 10, 2, 9 \rangle$.

$\frac{x - \frac{11}{3}}{10} = \frac{y - \frac{7}{3}}{2} = \frac{z}{9}$

3. Final Answer

$\frac{x - \frac{11}{3}}{10} = \frac{y - \frac{7}{3}}{2} = \frac{z}{9}$