The problem asks you to:

1. Find the equation of a hyperplane passing through the origin and in the direction of the vectors $\mathbf{v}_1 = (1, 2, 2)$ and $\mathbf{v}_2 = (4, 1, -1)$.
2. Find where this hyperplane intersects the line given by the parametric equations: $x_1 = -t + 2, \quad x_2 = 2t - 1, \quad x_3 = t + 3$

Part 1: Equation of the Hyperplane

The equation of a hyperplane in 3D space through the origin can be expressed as: $a_1x_1 + a_2x_2 + a_3x_3 = 0$ The normal vector to the hyperplane will be perpendicular to both given vectors $\mathbf{v}_1 = (1, 2, 2)$ and $\mathbf{v}_2 = (4, 1, -1)$. To find the normal vector $\mathbf{n}$, we can compute the cross product $\mathbf{v}_1 \times \mathbf{v}_2$.

Part 2: Intersection of the Hyperplane and the Line

Substitute the parametric equations of the line into the equation of the hyperplane and solve for the value of $t$ at which the line intersects the hyperplane.

Let me perform these calculations step-by-step.

Step 1: Find the Cross Product

The cross product of $\mathbf{v}_1 = (1, 2, 2)$ and $\mathbf{v}_2 = (4, 1, -1)$ is computed as: $\mathbf{n} = \mathbf{v}_1 \times \mathbf{v}_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & 2 \\ 4 & 1 & -1 \end{vmatrix}$

This yields: $\mathbf{n} = \mathbf{i}(2(-1) - 2(1)) - \mathbf{j}(1(-1) - 2(4)) + \mathbf{k}(1(1) - 2(4))$

Let me compute this. numpy as np

Define vectors v1 and v2

v1 = np.array([1, 2, 2]) v2 = np.array([4, 1, -1])

Compute the cross product

n = np.cross(v1, v2) n