We are tasked with finding a unit vector that is orthogonal to both vectors $\mathbf{i} + \mathbf{j}$ and $\mathbf{i} + \mathbf{k}$.

Step 1: Represent the given vectors

We can represent the vectors $\mathbf{i} + \mathbf{j}$ and $\mathbf{i} + \mathbf{k}$ in component form:

• $\mathbf{i} + \mathbf{j} = (1, 1, 0)$
• $\mathbf{i} + \mathbf{k} = (1, 0, 1)$

Step 2: Find a vector orthogonal to both

To find a vector orthogonal to both, we can compute the cross product of these two vectors.

Let: $\mathbf{v}_1 = (1, 1, 0) \quad \text{and} \quad \mathbf{v}_2 = (1, 0, 1)$

The cross product $\mathbf{v}_1 \times \mathbf{v}_2$ is given by:

\mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{vmatrix}$$ This determinant expands to: $$\mathbf{v}_1 \times \mathbf{v}_2 = \mathbf{i}(1 \cdot 1 - 0 \cdot 1) - \mathbf{j}(1 \cdot 1 - 0 \cdot 1) + \mathbf{k}(1 \cdot 0 - 1 \cdot 1)$$ $$\mathbf{v}_1 \times \mathbf{v}_2 = \mathbf{i}(1) - \mathbf{j}(1) + \mathbf{k}(-1)$$ $$\mathbf{v}_1 \times \mathbf{v}_2 = (1, -1, -1)$$ ### Step 3: Find the unit vector The magnitude of the vector $$(1, -1, -1)$$ is: $$|\mathbf{v}| = \sqrt{1^2 + (-1)^2 + (-1)^2} = \sqrt{1 + 1 + 1} = \sqrt{3}$$ To find the unit vector, divide each component by the magnitude: $$\mathbf{u} = \frac{1}{\sqrt{3}}(1, -1, -1) = \left(\frac{1}{\sqrt{3}}, \frac{-1}{\sqrt{3}}, \frac{-1}{\sqrt{3}}\right)$$ ### Final Answer: The unit vector that is orthogonal to both $$\mathbf{i} + \mathbf{j}$$ and $$\mathbf{i} + \mathbf{k}$$ is: $$\left(\frac{1}{\sqrt{3}}, \frac{-1}{\sqrt{3}}, \frac{-1}{\sqrt{3}}\right)$$ Would you like further details on any of the steps? --- Here are 5 related questions to explore: 1. What is the physical significance of a cross product in vector mathematics? 2. How do you find the magnitude of any given vector? 3. What other methods can be used to find an orthogonal vector besides the cross product? 4. How would the cross product change if different vectors were used? 5. How do we normalize vectors in higher dimensions? **Tip**: When dealing with unit vectors, always remember to normalize them by dividing by their magnitude to ensure their length is 1.