Find two unit vectors orthogonal to 
 and 

Enter your answer so that the first vector has a positive first coordinate and the second vector has a negative first coordinate

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.

Find two unit vectors orthogonal to vectors (2, -1, 3) and (1, 4, -2). Enter your answer so that the first vector has a positive first coordinate and the second vector has a negative first coordinate.

This guide explains how to find two unit vectors orthogonal to the vectors (2, -1, 3) and (1, 4, -2). Learn how to compute the cross product, normalize the resulting vector, and adjust the sign to meet specific coordinate conditions. Suitable for Grades 10-12 students studying vector algebra.

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