The given problem asks to find the vector equation for the line of intersection of two planes:

1. $2x + 5y - 3z = -5$
2. $2x + 4z = 1$

The vector equation provided is: $\mathbf{r} = \left(\frac{1}{2}, -\frac{6}{5}, 0\right) + t \left(20, -\frac{1}{5}, 1\right)$

Here’s how this works:

Step-by-Step Breakdown:

1. Find a point of intersection:
   To find the vector equation of the line, we first need to find a specific point that lies on both planes (a point of intersection). This can be done by solving the system of plane equations.

   $2x + 5y - 3z = -5$ $2x + 4z = 1$ Solve these simultaneously to find values for $x$, $y$, and $z$.

2. Find the direction vector:
   The direction of the line is given by the cross product of the normal vectors of the two planes. The normal vectors for the planes are: $\mathbf{n}_1 = (2, 5, -3) \quad \text{and} \quad \mathbf{n}_2 = (2, 0, 4)$ By taking the cross product $\mathbf{n}_1 \times \mathbf{n}_2$, we obtain the direction vector of the line.

3. Form the vector equation:
   The vector equation of the line takes the form: $\mathbf{r} = \mathbf{r}_0 + t \mathbf{v}$ where $\mathbf{r}_0$ is a point on the line (from step 1) and $\mathbf{v}$ is the direction vector (from step 2).

Solution provided:

• The point of intersection is $\left(\frac{1}{2}, -\frac{6}{5}, 0\right)$.
• The direction vector is $\left(20, -\frac{1}{5}, 1\right)$.
• Thus, the vector equation of the line is: $\mathbf{r} = \left(\frac{1}{2}, -\frac{6}{5}, 0\right) + t \left(20, -\frac{1}{5}, 1\right)$

Would you like more details on any of the steps?

Here are 5 related questions to explore further:

1. How do you find the cross product of two vectors?
2. What is the significance of the direction vector in a line's equation?
3. How can we interpret the vector equation geometrically?
4. What are other ways to find the intersection of two planes?
5. How can you verify that the given point lies on both planes?

Tip: When finding the line of intersection between two planes, always ensure that the direction vector is perpendicular to both planes’ normal vectors.