How can you find two unit vectors orthogonal to both vectors in a given plane?

To find two unit vectors orthogonal to both vectors in a given plane, we can use the cross product of the two given vectors. Let's denote the two given vectors as $\vec{a}$ and $\vec{b}$. The cross product of these two vectors, $\vec{a} \times \vec{b}$, will give us a vector that is orthogonal to both $\vec{a}$ and $\vec{b}$.

Here are the steps to find the two unit vectors orthogonal to both $\vec{a}$ and $\vec{b}$:

Step 1: Compute the Cross Product
Given two vectors $\vec{a} = \langle a_1, a_2, a_3 \rangle$ and $\vec{b} = \langle b_1, b_2, b_3 \rangle$, their cross product $\vec{c} = \vec{a} \times \vec{b}$ is calculated as follows:
$$
\vec{c} = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
a_1 & a_2 & a_3 \\
b_1 & b_2 & b_3 \\
\end{vmatrix}
= \langle (a_2b_3 - a_3b_2), (a_3b_1 - a_1b_3), (a_1b_2 - a_2b_1) \rangle
$$
where $\mathbf{i}$, $\mathbf{j}$, and $\mathbf{k}$ are the unit vectors in the direction of the x, y, and z axes, respectively.

Step 2: Normalize the Resulting Vector
To convert the resulting vector $\vec{c}$ into a unit vector, we need to divide it by its magnitude. The magnitude of $\vec{c}$, denoted as $||\vec{c}||$, is given by:
$$
||\vec{c}|| = \sqrt{c_1^2 + c_2^2 + c_3^2}
$$
where $c_1$, $c_2$, and $c_3$ are the components of $\vec{c}$.

The unit vector $\vec{u}$ in the direction of $\vec{c}$ is then:
$$
\vec{u} = \frac{\vec{c}}{||\vec{c}||} = \left\langle \frac{c_1}{||\vec{c}||}, \frac{c_2}{||\vec{c}||}, \frac{c_3}{||\vec{c}||} \right\rangle
$$

Step 3: Find the Second Unit Vector
Since the cross product only gives us one vector orthogonal to the plane defined by $\vec{a}$ and $\vec{b}$, and we need two unit vectors, we can take advantage of the fact that if $\vec{u}$ is orthogonal to the plane, then so is $-\vec{u}$. Therefore, the second unit vector $\vec{v}$ is simply the negation of $\vec{u}$:
$$
\vec{v} = -\vec{u} = \left\langle -\frac{c_1}{||\vec{c}||}, -\frac{c_2}{||\vec{c}||}, -\frac{c_3}{||\vec{c}||} \right\rangle
$$

Example:
Let's say we have two vectors $\vec{a} = \langle 1, 2, 3 \rangle$ and $\vec{b} = \langle 4, 5, 6 \rangle$. We want to find two unit vectors orthogonal to both $\vec{a}$ and $\vec{b}$.

# Step 1: Compute the Cross Product
$$
\vec{c} = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
1 & 2 & 3 \\
4 & 5 & 6 \\
\end{vmatrix}
= \langle (2 \cdot 6 - 3 \cdot 5), (3 \cdot 4 - 1 \cdot 6), (1 \cdot 5 - 2 \cdot 4) \rangle
= \langle 12 - 15, 12 - 6, 5 - 8 \rangle
= \langle -3, 6, -3 \rangle
$$

# Step 2: Normalize the Resulting Vector
$$
||\vec{c}|| = \sqrt{(-3)^2 + 6^2 + (-3)^2} = \sqrt{9 + 36 + 9} = \sqrt{54} = 3\sqrt{6}
$$
$$
\vec{u} = \frac{\vec{c}}{||\vec{c}||} = \left\langle \frac{-3}{3\sqrt{6}}, \frac{6}{3\sqrt{6}}, \frac{-3}{3\sqrt{6}} \right\rangle = \left\langle -\frac{1}{\sqrt{6}}, \frac{2}{\sqrt{6}}, -\frac{1}{\sqrt{6}} \right\rangle
$$

# Step 3: Find the Second Unit Vector
$$
\vec{v} = -\vec{u} = \left\langle \frac{1}{\sqrt{6}}, -\frac{2}{\sqrt{6}}, \frac{1}{\sqrt{6}} \right\rangle
$$

Thus, the two unit vectors orthogonal to both $\vec{a}$ and $\vec{b}$ are $\vec{u}$ and $\vec{v}$.