a) Find parametric equations for the line through (4, 1, 4) that is perpendicular to the plane x − y + 2z = 7. (Use the parameter t.)
b) In what points does this line intersect the coordinate planes?
xy-plane?
yz-plane?
xz-plane?
the normal to the plane has direction cosines proportional to <1,-1,2>, so
x = 4+t
y = 1-t
z = 4+2t
xy-intercept:
z=0, so t = -2
(2,3,0)
and so on
a) To find the parametric equations for a line that is perpendicular to a plane, we need to find the direction vector of the line. Since the line is perpendicular to the plane, its direction vector will be orthogonal (perpendicular) to the plane's normal vector.
The given plane has the equation: x - y + 2z = 7.
First, let's find the normal vector of the plane. The coefficients of x, y, and z represent the normal vector components. So, the normal vector of the plane is <1, -1, 2>.
A direction vector of the line can be obtained by taking the cross product between the normal vector of the plane and any vector in the line. For convenience, let's choose a vector from the line passing through the point (4, 1, 4). Subtracting this point's coordinates from any point on the line should give us a vector parallel to the line.
Let's use P(4, 1, 4) as a point on the line. Subtracting the coordinates of P from any point on the line should give us the line's direction vector. Let Q(x, y, z) be a point on the line. Subtracting P from Q gives us (x - 4, y - 1, z - 4).
This vector (x - 4, y - 1, z - 4) should be orthogonal to the plane's normal vector. Therefore, we can set up a dot product equation:
<1, -1, 2> ⋅ (x - 4, y - 1, z - 4) = 0
Simplifying the dot product, we get:
(x - 4) - (y - 1) + 2(z - 4) = 0
Expanding and rearranging the equation:
x - y + 2z - 12 = 0
Rearranging further:
x - y + 2z = 12
We now have the equation of a plane perpendicular to the line we are looking for. But notice that this is the same plane that was given initially. Therefore, the line we are looking for lies entirely on this plane.
The parametric equations for the line can be written as:
x = 4 + t
y = 1 + t
z = 4 - 2t
Here, t represents the parameter that helps us generate all the points on the line.
b) To find the points where the line intersects the coordinate planes, we set one of the variables (x, y, or z) to zero at a time and solve for the other two variables.
1. Intersection with the xy-plane (z = 0):
Setting z = 0 in the parametric equations gives:
x = 4 + t
y = 1 + t
z = 0
Solving x = 4 + t and y = 1 + t simultaneously, we get:
x - 4 = y - 1
t = y - x + 3
Thus, the line intersects the xy-plane at the point (4, 1, 0).
2. Intersection with the yz-plane (x = 0):
Setting x = 0 in the parametric equations gives:
x = 0
y = 1 + t
z = 4 - 2t
The line intersects the yz-plane at the point (0, 1, 4).
3. Intersection with the xz-plane (y = 0):
Setting y = 0 in the parametric equations gives:
x = 4 + t
y = 0
z = 4 - 2t
The line intersects the xz-plane at the point (4, 0, 4).
In summary, the line intersects the coordinate planes at the following points:
xy-plane: (4, 1, 0)
yz-plane: (0, 1, 4)
xz-plane: (4, 0, 4)