Find parametric equations for the line that is perpendicular to the plane 2x-y+3z=7 and intersects the plane when x=1 and y=1.
when x=1 and y = 1
in the plane:
2-1 + 3z = 7
z = 2
so the line will intersect at (1,1,2)
and the direction of the line will be (2,-1,3), the normal to the plane
line equation:
x = 1 + 2t
y = 1 - t
z = 2 + 3t
To find parametric equations for a line that is perpendicular to a plane and intersects it at a given point, you can follow these steps:
1. Find the normal vector of the given plane.
2. Use the normal vector to determine the direction vector of the line.
3. Use the given point of intersection to find the initial position vector of the line.
4. Write the parametric equations using the calculated direction vector and initial position vector.
Let's go through each step:
1. Find the normal vector of the given plane:
The coefficients of x, y, and z in the equation of the plane 2x-y+3z=7 represent the components of the normal vector. The normal vector is [2, -1, 3].
2. Determine the direction vector of the line:
Since the line is perpendicular to the plane, the direction vector of the line will be parallel to the normal vector of the plane. So, the direction vector of the line is [2, -1, 3].
3. Find the initial position vector of the line:
The point of intersection is given as (x = 1, y = 1). We can use this point as the initial position vector. So, the initial position vector is [1, 1, 0].
4. Write the parametric equations:
The parametric equations of a line are typically written as:
x = x_0 + a * t (1)
y = y_0 + b * t (2)
z = z_0 + c * t (3)
where (x_0, y_0, z_0) is the initial position vector, and (a, b, c) is the direction vector.
Using the given values:
Initial position vector: [1, 1, 0]
Direction vector of the line: [2, -1, 3]
Therefore, the parametric equations for the line would be:
x = 1 + 2t
y = 1 - t
z = 3t
These equations describe a line that is perpendicular to the plane 2x-y+3z=7 and intersects the plane at (1, 1, 0).