To find the vector equation of a plane in $\mathbb{R}^4$ that passes through the point $p = (1, 3, 5, -2)$ and has the direction vectors $v = [0, 2, -2, 3]$ and $w = [1, 1, -1, 1]$, you can use the following vector equation:

$x = p + sv + tw$

Where:

• $x$ is any point on the plane,
• $p$ is the given point on the plane $(1, 3, 5, -2)$,
• $v$ and $w$ are direction vectors, and
• $s$ and $t$ are scalar parameters.

Now, breaking this down step by step:

1. The position vector of the point $p$ is: $p = [1, 3, 5, -2]$

2. Add the direction vector $v$ scaled by $s$: [ s[0, 2, -2, 3] = [0, 2s, -2s, 3s] ]

3. Add the direction vector $w$ scaled by $t$: [ t[1, 1, -1, 1] = [t, t, -t, t] ]

Thus, the vector equation for the plane is:

[ x = [1, 3, 5, -2] + s[0, 2, -2, 3] + t[1, 1, -1, 1] ]

Expanding this:

[ x = [1, 3, 5, -2] + [0, 2s, -2s, 3s] + [t, t, -t, t] ]

Final answer:

$x = [1 + t, 3 + 2s + t, 5 - 2s - t, -2 + 3s + t]$

This is the vector equation of the plane in $\mathbb{R}^4$.

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Would you like more details on the steps or have any questions?

Here are 5 related questions:

1. What is the general form of a vector equation for a plane in $\mathbb{R}^n$?
2. How can you determine if a point lies on the plane described by this equation?
3. How does this equation change if one or both direction vectors are parallel?
4. How do you find the normal vector to the plane in $\mathbb{R}^3$ from direction vectors?
5. Can this method be extended to define a plane in $\mathbb{R}^5$?

Tip: Always check if your direction vectors are linearly independent to ensure they define a unique plane.