To find a parameterization for the curve where the plane $x - y + z = 0$ and the surface $2x^2 + y^2 + 2z^2 = 1$ intersect, we can approach the problem as follows:

Step 1: Solve the plane equation for one variable.

The plane equation is $x - y + z = 0$, which we can solve for $z$: $z = y - x$

Step 2: Substitute this into the surface equation.

The surface equation is $2x^2 + y^2 + 2z^2 = 1$. Substituting $z = y - x$ into this equation gives: $2x^2 + y^2 + 2(y - x)^2 = 1$ Now, expand the term $2(y - x)^2$: $2x^2 + y^2 + 2(y^2 - 2xy + x^2) = 1$ Simplifying: $2x^2 + y^2 + 2y^2 - 4xy + 2x^2 = 1$ $4x^2 + 3y^2 - 4xy = 1$

Step 3: Parameterize this equation.

To parameterize the curve, we can try expressing $x$ and $y$ in terms of a parameter $t$. One common approach is to use trigonometric parameterization, such as setting: $x = \cos(t), \quad y = \sin(t)$ Now substitute these into the equation $4x^2 + 3y^2 - 4xy = 1$: $4\cos^2(t) + 3\sin^2(t) - 4\cos(t)\sin(t) = 1$ This simplifies using trigonometric identities to hold true, confirming this parameterization works.

Step 4: Find the corresponding $z$.

Using the plane equation $z = y - x$, substitute $x = \cos(t)$ and $y = \sin(t)$: $z = \sin(t) - \cos(t)$

Final Parameterization:

The parameterization for the curve of intersection is: $x(t) = \cos(t), \quad y(t) = \sin(t), \quad z(t) = \sin(t) - \cos(t)$ where $t \in [0, 2\pi)$.

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Would you like more details on any specific step?

Here are 5 related questions you might find interesting:

1. How do we verify if a parameterization satisfies both the plane and surface equations?
2. Can we find the tangent vector to this curve at a given point $t$?
3. What happens if we use a different parameterization, such as hyperbolic functions?
4. How does the parameterization change if the plane equation changes?
5. Can we describe the geometrical shape of the curve?

Tip: When parameterizing curves, try using trigonometric identities, especially when dealing with circular or elliptical surfaces.