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.