1. **Problem statement:** Find vectors in $\mathbb{R}^4$ that form an orthogonal basis with the given vectors $\mathbf{v}_1 = (1, -2, 2, -3)$ and $\mathbf{v}_2 = (2, -3, 2, 4)$. This means we want vectors orthogonal to both $\mathbf{v}_1$ and $\mathbf{v}_2$. 2. **Key concept:** Two vectors $\mathbf{a}$ and $\mathbf{b}$ are orthogonal if their dot product is zero: $$\mathbf{a} \cdot \mathbf{b} = 0.$$ 3. **Set up the problem:** Let $\mathbf{x} = (x_1, x_2, x_3, x_4)$ be a vector orthogonal to both $\mathbf{v}_1$ and $\mathbf{v}_2$. Then: $$ \mathbf{x} \cdot \mathbf{v}_1 = x_1(1) + x_2(-2) + x_3(2) + x_4(-3) = 0 $$ $$ \mathbf{x} \cdot \mathbf{v}_2 = x_1(2) + x_2(-3) + x_3(2) + x_4(4) = 0 $$ 4. **Write the system of equations:** $$ \begin{cases} x_1 - 2x_2 + 2x_3 - 3x_4 = 0 \\ 2x_1 - 3x_2 + 2x_3 + 4x_4 = 0 \end{cases} $$ 5. **Solve the system:** From the first equation: $$ x_1 = 2x_2 - 2x_3 + 3x_4 $$ Substitute into the second: $$ 2(2x_2 - 2x_3 + 3x_4) - 3x_2 + 2x_3 + 4x_4 = 0 $$ Simplify: $$ 4x_2 - 4x_3 + 6x_4 - 3x_2 + 2x_3 + 4x_4 = 0 $$ $$ (4x_2 - 3x_2) + (-4x_3 + 2x_3) + (6x_4 + 4x_4) = 0 $$ $$ x_2 - 2x_3 + 10x_4 = 0 $$ 6. **Express $x_2$ in terms of $x_3$ and $x_4$:** $$ x_2 = 2x_3 - 10x_4 $$ 7. **Express $x_1$ in terms of $x_3$ and $x_4$:** $$ x_1 = 2(2x_3 - 10x_4) - 2x_3 + 3x_4 = 4x_3 - 20x_4 - 2x_3 + 3x_4 = 2x_3 - 17x_4 $$ 8. **General solution:** $$ \mathbf{x} = (x_1, x_2, x_3, x_4) = (2x_3 - 17x_4, 2x_3 - 10x_4, x_3, x_4) $$ 9. **Choose basis vectors by assigning free variables:** - For $x_3 = 1$, $x_4 = 0$: $$ \mathbf{u} = (2, 2, 1, 0) $$ - For $x_3 = 0$, $x_4 = 1$: $$ \mathbf{w} = (-17, -10, 0, 1) $$ 10. **Conclusion:** The vectors $\mathbf{u} = (2, 2, 1, 0)$ and $\mathbf{w} = (-17, -10, 0, 1)$ form an orthogonal basis with $\mathbf{v}_1$ and $\mathbf{v}_2$ in $\mathbb{R}^4$ because they are orthogonal to both given vectors and linearly independent. **Final answer:** $$ \boxed{\mathbf{u} = (2, 2, 1, 0), \quad \mathbf{w} = (-17, -10, 0, 1)} $$