Given three vectors $\mathbf{u} = \langle 1, 1, -2 \rangle$, $\mathbf{v} = \langle -1, 2, 0 \rangle$, and $\mathbf{w} = \langle -1, 0, 1 \rangle$, we need to find a vector $\mathbf{n} = \langle x, y, z \rangle$ that is orthogonal to all three vectors. This means that the dot product of $\mathbf{n}$ with each of $\mathbf{u}$, $\mathbf{v}$, and $\mathbf{w}$ must be zero.

The conditions for orthogonality are:

$\mathbf{n} \cdot \mathbf{u} = x(1) + y(1) + z(-2) = 0$

$\mathbf{n} \cdot \mathbf{v} = x(-1) + y(2) + z(0) = 0$

$\mathbf{n} \cdot \mathbf{w} = x(-1) + y(0) + z(1) = 0$

We can express these equations as:

1. $x + y - 2z = 0$
2. $-x + 2y = 0$
3. $-x + z = 0$

Let's solve this system of linear equations to find $x$, $y$, and $z$.

Solving the System

From the second equation, we have: $x = 2y$

From the third equation, we have: $x = z$

Substituting $x = z$ into the first equation: $z + y - 2z = 0 \implies y = z$

Thus, $x = y = z$.

Given that $x < 0$, let $x = -k$, where $k > 0$. Therefore, the vector $\mathbf{n}$ is:

$\mathbf{n} = \langle -k, -k, -k \rangle$

For simplicity, we can choose $k = 1$, so the vector $\mathbf{n}$ is:

$\mathbf{n} = \langle -1, -1, -1 \rangle$

Final Answer

The vector $\mathbf{n} = \langle -1, -1, -1 \rangle$ is orthogonal to all three vectors $\mathbf{u}$, $\mathbf{v}$, and $\mathbf{w}$.

Would you like a detailed explanation of the process, or do you have any further questions?

Here are 5 related questions for further exploration:

1. What would happen if $x > 0$?
2. How would the solution change if one of the given vectors was different?
3. Can there be other solutions for $\mathbf{n}$ apart from a scalar multiple?
4. What is the geometric interpretation of this orthogonal vector $\mathbf{n}$?
5. How can the cross product be used to find a vector orthogonal to two vectors?

Tip: Always check the consistency of the solution with the original conditions, especially when dealing with inequalities like $x < 0$.