Determine a vector equation for the plane that is parallel to the z-axis containing the line r = (1, 2, -1) + t(3, -1, 2).
To find a vector equation for the plane, we need a point on the plane and a normal vector to the plane.
The line r = (1, 2, -1) + t(3, -1, 2) has a direction vector of (3, -1, 2).
Since the plane is parallel to the z-axis, it means the normal vector to the plane is perpendicular to the z-axis. Therefore, a possible normal vector is (0, 0, 1).
We can choose a point on the line: when t = 0, r = (1, 2, -1).
So, a point on the plane is (1, 2, -1), and a normal vector to the plane is (0, 0, 1).
Therefore, a vector equation for the plane is:
(r - (1, 2, -1)) · (0, 0, 1) = 0
Simplifying, we get:
(0, 0, 1) · (x - 1, y - 2, z + 1) = 0
0 + 0 + (z + 1) = 0
z + 1 = 0
z = -1
So, the equation of the plane is z = -1.
To find a vector equation for the plane that is parallel to the z-axis and contains the line, we need to find two vectors in the plane.
The direction vector of the line is (3, -1, 2). To find a vector parallel to the z-axis, we can take the cross product of (3, -1, 2) with the unit vector in the z-direction, which is (0, 0, 1).
Let's calculate the cross product:
(3, -1, 2) x (0, 0, 1) = ((-1)(1) - (2)(0), (2)(0) - (3)(0), (3)(0) - (-1)(0)) = (-1, 0, 0)
So, (-1, 0, 0) is a vector parallel to the z-axis and in the plane.
Now, we can find another vector in the plane by taking the cross product of (-1, 0, 0) with the direction vector of the line:
(-1, 0, 0) x (3, -1, 2) = ((0)(2) - (0)(-1), (0)(3) - (-1)(-1), (-1)(-1) - (0)(3)) = (0, -1, 1)
So, (0, -1, 1) is another vector in the plane.
Finally, we can write the vector equation of the plane as:
r = (1, 2, -1) + s(-1, 0, 0) + t(0, -1, 1)
where (1, 2, -1) is a point on the line, and (s, t) are parameters.
To determine a vector equation for the plane that is parallel to the z-axis and contains the given line, we need to find two non-parallel vectors that lie on the plane. One approach is to find the perpendicular vector to the z-axis and use it as one of the vectors in the plane.
The z-axis is parallel to the vector (0, 0, 1). To find a vector perpendicular to this, we can take the cross product of (0, 0, 1) with any non-zero vector in the plane. In this case, we can take the cross product of (0, 0, 1) with the direction vector of the given line: (3, -1, 2).
The cross product of two vectors, v = (v1, v2, v3) and w = (w1, w2, w3), is given by the formula:
v x w = (v2w3 - v3w2, v3w1 - v1w3, v1w2 - v2w1)
Applying this formula, we have:
(0, 0, 1) x (3, -1, 2) = (1*(-1) - 2*(0), 2*(3) - (-1)*(1), 0 - 1*(3)) = (-1, 7, -3)
So, we have found a vector (-1, 7, -3) that is perpendicular to both the z-axis and the direction vector of the given line. This vector is in the plane we are looking for.
Now, we can use the given point on the line, (1, 2, -1), as a point in the plane. The equation of the plane can be written as:
r = (1, 2, -1) + s(-1, 7, -3) + t(0, 0, 1)
where r is a vector in the plane, s and t are scalar parameters, and (-1, 7, -3) is the vector perpendicular to the plane.
Therefore, a vector equation for the plane that is parallel to the z-axis and contains the line r = (1, 2, -1) + t(3, -1, 2) is:
r = (1, 2, -1) + s(-1, 7, -3) + t(0, 0, 1)