How can I find the parametric equations for a line that goes through the point (2, 4, 6) and is perpendicular to the plane defined by the equation x + y + 3z = 7?

by

To find the parametric equations for a line that passes through a specific point and is perpendicular to a given plane, we must first identify the normal vector to the plane and use that to define the direction of the line.

1. **Identify the normal vector to the plane:** The equation of the plane is given in the standard form as x + y + 3z = 7. From this equation, we can extract the coefficients of x, y, and z to determine the normal vector. The normal vector N can be defined as:

N = (1, 1, 3)

2. **Use the point and the normal vector to define the line:** The line that we are looking for passes through the point (2, 4, 6) and has the direction of the normal vector (1, 1, 3). We can define the parametric equations for the line using the point and the direction vector. The parametric equations can be formulated as follows:

x = x_0 + t * a
y = y_0 + t * b
z = z_0 + t * c

Where (x_0, y_0, z_0) is the point on the line, (a, b, c) is the direction vector, and t is a parameter that can take any real value. In our case, we have:

  • (x_0, y_0, z_0) = (2, 4, 6)
  • (a, b, c) = (1, 1, 3)

Substituting these values into the equations gives:

x = 2 + t * 1 = 2 + t
y = 4 + t * 1 = 4 + t
z = 6 + t * 3 = 6 + 3t

3. **Final parametric equations:** Thus, the parametric equations for the line are:

  • x(t) = 2 + t
  • y(t) = 4 + t
  • z(t) = 6 + 3t

4. **Conclusion:** These equations describe the line that passes through the point (2, 4, 6) and is perpendicular to the plane x + y + 3z = 7. As t varies over all real numbers, the equations trace out the line in three-dimensional space.

Leave a Comment