Determine a vector equation for the plane that is parallel to the xz -plane and passes through the point (-4, -1, 3).
Olivia/Darren/Harvey -- quite the identity crisis going on here!
Please pick one name to use.
Let r be the position vector of any point (x,y,z) in the plane, and B = -4i-j+3k
A vector perpendicular to the xz plane is just A = j
Since B-r lies in the plane, A•(B-r) = 0
To determine a vector equation for a plane, we need two direction vectors that lie on the plane.
In this case, since the plane is parallel to the xz-plane, the normal vector of the plane would be orthogonal to the xz-plane and would have a y-component of 1 (since the normal vector needs to be perpendicular to the plane).
So, the normal vector of the plane is (0, 1, 0).
Now, we need another vector that lies on the plane. We can choose any vector that is parallel to the xz-plane, such as (1, 0, 0).
To find the equation of the plane, we can use the point-normal form of the equation of a plane:
n · (r - r0) = 0, where n is the normal vector, r is a position vector on the plane, and r0 is a position vector of a point on the plane.
Substituting the given values, the equation becomes:
(0, 1, 0) · (r - (-4, -1, 3)) = 0
(0, 1, 0) · (r + (4, 1, -3)) = 0
Expanding the dot product, we get:
(0)(r1 + 4) + (1)(r2 + 1) + (0)(r3 - 3) = 0
r2 + 1 = 0
r2 = -1
Therefore, a vector equation for the plane that is parallel to the xz-plane and passes through the point (-4, -1, 3) is:
r = (-4, -1, 3) + t(1, -1, 0)