Find the parametric equations for the line through the point P = (5, -1, -3) that is perpendicular to the plane 4x−4y+0z=1.
Use "t" as your variable, t = 0 should correspond to P, and the velocity vector of the line should be the same as the standard normal vector of the plane.
x =
y =
z =
t = (x-5)/4 = (y+1)/-4
x = 4t + 5
y = -4t - 1
as well as
z = -3
oh, yeah -- what he said.
To find the parametric equations for the line through the point P = (5, -1, -3) that is perpendicular to the plane 4x − 4y + 0z = 1, we need to find the direction vector of the line.
The normal vector of the plane is [4, -4, 0], which is perpendicular to the plane. Since we want the velocity vector of the line to be the same as this normal vector, the direction vector of the line will be the same as the normal vector.
So, the direction vector of the line is [4, -4, 0].
Now, let's write the parametric equations of the line using the point P and the direction vector.
Since t = 0 should correspond to P = (5, -1, -3), we have:
x = 5 + 4t (equation 1)
y = -1 - 4t (equation 2)
z = -3 + 0t (equation 3)
So, the parametric equations for the line are:
x = 5 + 4t
y = -1 - 4t
z = -3
This represents a line passing through the point P = (5, -1, -3) and with a direction vector [4, -4, 0].