Find the vector equestion for line of intersection of the planes 5x+5y + z =4 and 5x + 4z = 5

We are tasked with finding the vector equation of the line formed by the intersection of two planes:

1. \( 5x + 5y + z = 4 \)
2. \( 5x + 4z = 5 \)

### Step 1: Find a Direction Vector
The direction vector of the line of intersection is parallel to the cross product of the normals of the two planes. First, let's identify the normals of both planes.

- The normal vector of the first plane \( 5x + 5y + z = 4 \) is \( \mathbf{n}_1 = \langle 5, 5, 1 \rangle \).
- The normal vector of the second plane \( 5x + 4z = 5 \) is \( \mathbf{n}_2 = \langle 5, 0, 4 \rangle \).

Now, we find the cross product of \( \mathbf{n}_1 \) and \( \mathbf{n}_2 \) to get the direction vector of the line.

\[
\mathbf{d} = \mathbf{n}_1 \times \mathbf{n}_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 5 & 5 & 1 \\ 5 & 0 & 4 \end{vmatrix}
\]

\[
\mathbf{d} = \mathbf{i}(5 \times 4 - 1 \times 0) - \mathbf{j}(5 \times 4 - 1 \times 5) + \mathbf{k}(5 \times 0 - 5 \times 5)
\]

\[
\mathbf{d} = \mathbf{i}(20) - \mathbf{j}(20 - 5) + \mathbf{k}(0 - 25)
\]

\[
\mathbf{d} = \langle 20, -15, -25 \rangle
\]

Thus, the direction vector is \( \mathbf{d} = \langle 20, -15, -25 \rangle \).

### Step 2: Find a Point on the Line
We now need to find a point that lies on both planes. To do this, we solve the system of equations:

\[
5x + 5y + z = 4 \quad \text{(1)}
\]
\[
5x + 4z = 5 \quad \text{(2)}
\]

From equation (2), we solve for \( x \):

\[
5x = 5 - 4z \quad \Rightarrow \quad x = 1 - \frac{4z}{5}
\]

Substitute this into equation (1):

\[
5(1 - \frac{4z}{5}) + 5y + z = 4
\]

Simplify:

\[
5 - 4z + 5y + z = 4 \quad \Rightarrow \quad 5y - 3z = -1
\]

Solve for \( y \):

\[
y = \frac{3z - 1}{5}
\]

Choose \( z = 0 \) to simplify the calculations:

- If \( z = 0 \), then \( x = 1 \) and \( y = -\frac{1}{5} \).

Thus, the point \( P(1, -\frac{1}{5}, 0) \) lies on the line.

### Step 3: Write the Vector Equation
The vector equation of the line is:

\[
\mathbf{r}(t) = \mathbf{r}_0 + t\mathbf{d}
\]

Where:
- \( \mathbf{r}_0 = \langle 1, -\frac{1}{5}, 0 \rangle \) is the point on the line.
- \( \mathbf{d} = \langle 20, -15, -25 \rangle \) is the direction vector.

Thus, the vector equation is:

\[
\mathbf{r}(t) = \langle 1, -\frac{1}{5}, 0 \rangle + t \langle 20, -15, -25 \rangle
\]

Or equivalently:

\[
\mathbf{r}(t) = \langle 1 + 20t, -\frac{1}{5} - 15t, -25t \rangle
\]

This is the vector equation of the line of intersection of the two planes.

Would you like more details or further clarifications?

### Related Questions:
1. How do you find the equation of a plane given three points?
2. What is the geometric meaning of the cross product of two vectors?
3. How do you parameterize the intersection of two planes in 3D space?
4. How can you check if two planes are parallel or identical?
5. What is the algebraic method to solve systems of plane equations?

### Tip:
Choosing simple values for parameters like \( t \) or solving for variables early can help simplify complex geometry problems significantly!

Find the vector equation for the line of intersection of the planes 5x + 5y + z = 4 and 5x + 4z = 5.

Learn how to find the vector equation for the line of intersection of two planes: 5x + 5y + z = 4 and 5x + 4z = 5. This step-by-step solution covers vector geometry concepts, the cross product of normal vectors, and solving systems of plane equations. Suitable for undergraduate students.

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