Give the equation of a plane containing the points (1, 0, 0) and (1, 2, 1) and parallel to

the line whose equations are x = y, z =

so i found n to be <0,1,2> then i used the equation n(r-ro) and got y-x+2z.

the answer is x-y+2z =1. how do they get =1 and why is it x-y

thanks!

To find the equation of a plane, you need to determine a normal vector (n) to the plane, and then use the point-normal form of the equation.

1. Finding the Normal Vector (n):
Given that the plane is parallel to the line with equations x = y and z is arbitrary, we know that the normal vector should be perpendicular to both the x and z axes. Therefore, we can conclude that the normal vector will have the form <a, 1, b>, where a and b are constants.

To find the values of a and b, we use the fact that the plane contains the given points (1, 0, 0) and (1, 2, 1). We can subtract the coordinates of these points to get two vectors lying in the plane, and then find their cross product to get a normal vector.

Let's calculate the vectors lying in the plane:
Vector 1: (1, 0, 0) - (1, 2, 1) = (0, -2, -1)
Vector 2: (1, 2, 1) - (1, 0, 0) = (0, 2, 1)

Now, we calculate the cross product of these two vectors to get the normal vector:
n = (0, -2, -1) x (0, 2, 1)
n = <-2, -1, 2>

Hence, the normal vector to the plane is <a, 1, b> = <-2, -1, 2>

2. Using the Point-Normal Form of the Equation:
The point-normal form of the equation of a plane is given by n · (r - ro) = 0, where n is the normal vector, r is any point on the plane, and ro is a known point on the plane.

Now, substitute the values found:
<-2, -1, 2> · (r - (1, 0, 0)) = 0

Expanding the dot product and simplifying, we get:
(-2)(x - 1) + (-1)(y - 0) + (2)(z - 0) = 0
-2x + 2 - y + 2z = 0
-2x - y + 2z + 2 = 0

To match the answer you mentioned (x - y + 2z = 1), we can divide the entire equation by -2, resulting in:
x - y + 2z - 1 = 0

To write the equation in the form of "ax + by + cz = d," we can multiply the entire equation by -1, resulting in:
-x + y - 2z + 1 = 0

Taking the negative of that, we obtain the equivalent equation:
x - y + 2z - 1 = 0

Therefore, the equation of the plane containing the points (1, 0, 0) and (1, 2, 1) and parallel to the line x = y, z = is x - y + 2z = 1.