Answer:
2x -z = 3
Explanation:
You want the equation of the plane parallel to the y-axis and containing the points (1, 2, -1) and (2, 3, 1).
Normal
The normal to the plane will tell the coefficients in the equation. It can be found as the cross product of two vectors in the plane. One of them will be the difference of the two points:
(2, 3, 1) -(1, 2, -1) = (1, 1, 2)
Another will be the vector in the direction of the y-axis: (0, 1, 0).
The cross product of these is ...
(2, 3, 1) × (0, 1, 0) = (-2, 0, 1)
An equation for the plane could be ...
-2x +z = c . . . . . . for some suitable constant
We prefer a positive leading coefficient. Using the point (2, 3, 1) we can find c:
2x -z = 2(2) -(1) = 3
The equation of the plane is 2x -z = 3.
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Additional comment
As a check, we can use the other point: 2(1) -(-1) = 3, so it also satisfies the equation. Since there is no y-term in the equation, we know that y-values can be anything and the plane is parallel to the y-axis.
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