Answer:
x = 4/3 + t
y = -1/3 - 2t
z = 4/3 - t
Explanation:
The given line can be represented by the vector equation:
r = <1, 1, 0> + t<1, -1, 2>
We can find a vector that is perpendicular to this line by taking the cross product of the direction vector <1, -1, 2> with any other vector. Let's choose the vector <1, 0, 0> for this purpose:
n = <1, -1, 2> x <1, 0, 0> = <-2, -1, -1>
Now we have a normal vector n = <-2, -1, -1> to the line we want to find. We can use this vector and the given point (0, 1, 2) to find the equation of the plane that contains the line we want to find:
-2(x-0) - (y-1) - (z-2) = 0
-2x - y - z + 3 = 0
This plane intersects the given line when they have a point in common. To find this point, we can solve the system of equations:
-2x - y - z + 3 = 0
x - y = 1
z = 2t
From the second equation, we get x = t+1 and y = t. Substituting these into the first equation, we get:
-2(t+1) - t - 2t + 3 = 0
t = -1/3
Therefore, the point of intersection is (4/3, -1/3, 4/3). This point lies on both the line and the plane, so it is the point we need to use to find the parametric equations of the line we want to find.
Let's call the point we just found P. We can find the direction vector of the line we want to find by taking the cross product of the normal vector n with the vector from P to the point on the given line:
d = <-2, -1, -1> x <4/3-1, -1/3-1, 4/3-2> = <1, -2, -1>
Therefore, the parametric equations of the line we want to find are:
x = 4/3 + t
y = -1/3 - 2t
z = 4/3 - t