Find two unit vectors orthogonal to a=⟨−1,−4,5⟩ and b=⟨0,−4,−4⟩
Enter your answer so that the first non-zero coordinate of the first vector is positive.

To find two unit vectors orthogonal to both vectors \( \mathbf{a} = \langle -1, -4, 5 \rangle \) and \( \mathbf{b} = \langle 0, -4, -4 \rangle \), we can use the **cross product** of the two vectors. The result of the cross product will give a vector that is orthogonal to both \( \mathbf{a} \) and \( \mathbf{b} \). Once we compute the cross product, we will normalize the resulting vector to get a unit vector, and its negative will be the second unit vector.

### Step 1: Compute the Cross Product
The cross product of two vectors \( \mathbf{a} = \langle a_1, a_2, a_3 \rangle \) and \( \mathbf{b} = \langle b_1, b_2, b_3 \rangle \) is given by:

\[
\mathbf{a} \times \mathbf{b} = \langle a_2b_3 - a_3b_2, a_3b_1 - a_1b_3, a_1b_2 - a_2b_1 \rangle
\]

Substituting the components of \( \mathbf{a} = \langle -1, -4, 5 \rangle \) and \( \mathbf{b} = \langle 0, -4, -4 \rangle \):

\[
\mathbf{a} \times \mathbf{b} = \langle (-4)(-4) - (5)(-4), (5)(0) - (-1)(-4), (-1)(-4) - (-4)(0) \rangle
\]

Simplifying each component:

\[
\mathbf{a} \times \mathbf{b} = \langle 16 - (-20), 0 - 4, 4 - 0 \rangle = \langle 36, -4, 4 \rangle
\]

Thus, the cross product is:

\[
\mathbf{a} \times \mathbf{b} = \langle 36, -4, 4 \rangle
\]

### Step 2: Normalize the Cross Product
To find the unit vector, we need to normalize the cross product by dividing it by its magnitude. The magnitude of a vector \( \mathbf{v} = \langle v_1, v_2, v_3 \rangle \) is given by:

\[
|\mathbf{v}| = \sqrt{v_1^2 + v_2^2 + v_3^2}
\]

For \( \mathbf{v} = \langle 36, -4, 4 \rangle \):

\[
|\mathbf{v}| = \sqrt{36^2 + (-4)^2 + 4^2} = \sqrt{1296 + 16 + 16} = \sqrt{1328}
\]

Now, we normalize the vector:

\[
\mathbf{u} = \frac{1}{\sqrt{1328}} \langle 36, -4, 4 \rangle
\]

This gives the unit vector \( \mathbf{u} \).

### Step 3: Find the Two Unit Vectors
The two unit vectors orthogonal to both \( \mathbf{a} \) and \( \mathbf{b} \) are:

1. \( \frac{1}{\sqrt{1328}} \langle 36, -4, 4 \rangle \)
2. \( \frac{1}{\sqrt{1328}} \langle -36, 4, -4 \rangle \)

Since the first non-zero coordinate of the first vector is positive, the first unit vector is \( \frac{1}{\sqrt{1328}} \langle 36, -4, 4 \rangle \).

Do you want details or have any questions? Here are five questions related to this topic:

1. How do you verify that two vectors are orthogonal?
2. What are the properties of the cross product in vector algebra?
3. Why does the cross product result in a vector orthogonal to both original vectors?
4. How do you compute the magnitude of a vector in three dimensions?
5. What are some applications of the cross product in physics or engineering?

**Tip**: Always check the direction of the unit vectors to ensure you maintain the correct orientation for the first non-zero coordinate.

Find two unit vectors orthogonal to a=⟨−1,−4,5⟩ and b=⟨0,−4,−4⟩. Enter your answer so that the first non-zero coordinate of the first vector is positive.

Learn how to find two unit vectors orthogonal to vectors a = ⟨-1, -4, 5⟩ and b = ⟨0, -4, -4⟩ using the cross product method. This solution involves vector algebra concepts such as the cross product, magnitude, and normalization. Suitable for students in Grades 11-12 or college-level courses.

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