Final answer:
The intersection line of the planes x+y+z=0 and xy+z=0 can be parametrized with option a) x(t) = t, y(t) = -t, z(t) = 0 by using the point (1, -1, 0) and the direction vector (1, 1, -1) obtained from the cross product of the planes' normal vectors.
Step-by-step explanation:
To find a parametrization of the line at which the planes x+y+z=0 and xy+z=0 intersect, we first look for a point that lies on both planes. We can quickly verify that the point (1, -1, 0) satisfies both plane equations. To find the direction of the line of intersection, we can take the cross product of the normals to the planes.
For the planes given, the normal vectors are = and = . Using the cross product, we find × = <-k, -k, i-j>, which simplifies to (1, 1, -1). This gives us the direction of the line of intersection.
We can use the point (1, -1, 0) and the direction vector (1, 1, -1) to parametrize the line. If we let t be the parameter, the line is given by x(t) = 1+t, y(t) = -1+t, and z(t) = -t.
By adjusting these equations to have them start at the origin, we get x(t) = t, y(t) = -t, z(t) = 0 fulfilling the requirement for a parameterization and matching one of the given options.
Therefore, the correct parametrization of the line where the planes intersect is option a) x(t) = t, y(t) = -t, z(t) = 0.