Answer:
To find parametric equations for the line through (4, 4, 4) that is perpendicular to the plane x - y + 2z = 1, you can follow these steps:
(a) Parametric Equations for the Line:
1. First, find the normal vector to the plane x - y + 2z = 1. The coefficients of x, y, and z in this plane equation represent the components of the normal vector. So, the normal vector is N = (1, -1, 2).
2. Now, you need a direction vector for the line perpendicular to the plane. Any nonzero multiple of the normal vector N can serve as the direction vector. Let's use N itself as the direction vector.
3. Parametric equations for a line can be written as follows:
x(t) = x₀ + at
y(t) = y₀ + bt
z(t) = z₀ + ct
Here, (x₀, y₀, z₀) are the coordinates of a point on the line, and (a, b, c) are the components of the direction vector.
4. Plug in the values:
x(t) = 4 + t(1)
y(t) = 4 + t(-1)
z(t) = 4 + t(2)
So, the parametric equations for the line through (4, 4, 4) perpendicular to the plane x - y + 2z = 1 are:
x(t) = 4 + t
y(t) = 4 - t
z(t) = 4 + 2t
(b) To find the points where this line intersects the coordinate planes, set each of the coordinates equal to zero in the parametric equations and solve for t:
1. xy-plane (z = 0):
Set z(t) = 4 + 2t equal to 0 and solve for t:
4 + 2t = 0
2t = -4
t = -2
So, the line intersects the xy-plane at the point (4 - 2, 4 + 2, 0), which is (2, 6, 0).
2. yz-plane (x = 0):
Set x(t) = 4 + t equal to 0 and solve for t:
4 + t = 0
t = -4
So, the line intersects the yz-plane at the point (0, 4 - 4, 4 + (-4)), which is (0, 0, 0).
3. xz-plane (y = 0):
Set y(t) = 4 - t equal to 0 and solve for t:
4 - t = 0
t = 4
So, the line intersects the xz-plane at the point (4 + 4, 0, 4 + 2(4)), which is (8, 0, 12).
In summary, the line intersects the xy-plane at (2, 6, 0), the yz-plane at (0, 0, 0), and the xz-plane at (8, 0, 12).
Explanation: