Answer:To find the standard parametric equations for the line through the point (-2, 7.7) and perpendicular to the plane 7x + 5y + 5z = 24, we'll follow these steps:
1. Find the normal vector to the plane.
2. Use the normal vector to find the direction vector of the line.
3. Write the parametric equations for the line using the given point and direction vector.
Let's start with step 1:
1. **Find the normal vector to the plane:**
The coefficients of x, y, and z in the equation of the plane 7x + 5y + 5z = 24 give us the normal vector: (7, 5, 5).
Now, let's move to step 2:
2. **Use the normal vector to find the direction vector of the line:**
The direction vector of the line perpendicular to the plane is the same as the normal vector of the plane, but we'll negate its components to make it point in the opposite direction. So, the direction vector is (-7, -5, -5).
Finally, step 3:
3. **Write the parametric equations for the line using the given point and direction vector:**
The parametric equations for a line can be written as follows:
\[x = x_0 + at,\]
\[y = y_0 + bt,\]
\[z = z_0 + ct,\]
where (x, y, z) are the coordinates of any point on the line, (x₀, y₀, z₀) is the known point on the line, and (a, b, c) is the direction vector.
In this case, the given point is (-2, 7.7), and the direction vector is (-7, -5, -5). So, the parametric equations for the line are:
\[x = -2 - 7t,\]
\[y = 7.7 - 5t,\]
\[z = -5t.\]
This represents the line passing through the point (-2, 7.7) and perpendicular to the plane 7x + 5y + 5z = 24.
Explanation: