Let's solve the problem of finding two unit vectors orthogonal to vectors $\mathbf{a} = \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} 1 \\ 4 \\ -2 \end{pmatrix}$.

Step 1: Find the Cross Product

The cross product of $\mathbf{a}$ and $\mathbf{b}$, denoted as $\mathbf{a} \times \mathbf{b}$, gives a vector orthogonal to both $\mathbf{a}$ and $\mathbf{b}$.

The cross product is calculated as:

$\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -1 & 3 \\ 1 & 4 & -2 \end{vmatrix}$

This determinant expands to:

$\mathbf{a} \times \mathbf{b} = \mathbf{i}((-1)(-2) - (3)(4)) - \mathbf{j}((2)(-2) - (3)(1)) + \mathbf{k}((2)(4) - (-1)(1))$

Simplifying:

$\mathbf{a} \times \mathbf{b} = \mathbf{i}(2 - 12) - \mathbf{j}(-4 - 3) + \mathbf{k}(8 + 1)$

$\mathbf{a} \times \mathbf{b} = \mathbf{i}(-10) - \mathbf{j}(-7) + \mathbf{k}(9)$

$\mathbf{a} \times \mathbf{b} = \begin{pmatrix} -10 \\ 7 \\ 9 \end{pmatrix}$

Step 2: Normalize the Vector

Next, we need to normalize this vector to get a unit vector. The magnitude of $\mathbf{a} \times \mathbf{b}$ is:

$\|\mathbf{a} \times \mathbf{b}\| = \sqrt{(-10)^2 + 7^2 + 9^2} = \sqrt{100 + 49 + 81} = \sqrt{230}$

So the unit vector is:

$\mathbf{u_1} = \frac{1}{\sqrt{230}} \begin{pmatrix} -10 \\ 7 \\ 9 \end{pmatrix} = \begin{pmatrix} \frac{-10}{\sqrt{230}} \\ \frac{7}{\sqrt{230}} \\ \frac{9}{\sqrt{230}} \end{pmatrix}$

Step 3: Adjust for Sign of First Coordinate

We need one vector with a positive first coordinate and one with a negative first coordinate. The current vector $\mathbf{u_1}$ has a negative first coordinate. Therefore, we'll take the negative of this vector for $\mathbf{u_2}$:

$\mathbf{u_2} = \frac{1}{\sqrt{230}} \begin{pmatrix} 10 \\ -7 \\ -9 \end{pmatrix} = \begin{pmatrix} \frac{10}{\sqrt{230}} \\ \frac{-7}{\sqrt{230}} \\ \frac{-9}{\sqrt{230}} \end{pmatrix}$

Final Answers

1. First unit vector (positive first coordinate): $\mathbf{u_2} = \begin{pmatrix} \frac{10}{\sqrt{230}} \\ \frac{-7}{\sqrt{230}} \\ \frac{-9}{\sqrt{230}} \end{pmatrix}$

2. Second unit vector (negative first coordinate): $\mathbf{u_1} = \begin{pmatrix} \frac{-10}{\sqrt{230}} \\ \frac{7}{\sqrt{230}} \\ \frac{9}{\sqrt{230}} \end{pmatrix}$

Would you like more details on any part of this process or have any other questions?

Here are five related questions:

1. What is the geometric interpretation of the cross product of two vectors?
2. How do you determine if two vectors are orthogonal?
3. What does it mean for a vector to be a unit vector?
4. How is the magnitude of a vector calculated?
5. What is the significance of the sign of the first coordinate in vector normalization?

Tip: When finding unit vectors, always ensure the vector's magnitude is 1 by dividing by its length.