Decompose $$ \mathbf{v} $$ into two vectors $$ \mathbf{v}_{1} $$ and $$ \mathbf{v}_{2} $$, where $$ \mathbf{v}_{1} $$ is parallel to $$ \mathbf{w} $$ and $$ \mathbf{v}_{2} $$ is orthogonal to $$ \mathbf{w} $$. $$ \mathbf{v}=4 \mathbf{i}-6 \mathbf{j} $$ and $$ \mathbf{w}=\mathbf{i}+\mathbf{j} $$

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To decompose the vector $$ \mathbf{v} = 4 \mathbf{i} - 6 \mathbf{j} $$ into two components $$ \mathbf{v}_{1} $$ and $$ \mathbf{v}_{2} $$, where $$ \mathbf{v}_{1} $$ is parallel to $$ \mathbf{w} = \mathbf{i} + \mathbf{j} $$ and $$ \mathbf{v}_{2} $$ is orthogonal to $$ \mathbf{w} $$, we can follow these steps:

1. **Find the unit vector in the direction of $$ \mathbf{w} $$**:
   $$
   \|\mathbf{w}\| = \sqrt{1^2 + 1^2} = \sqrt{2}
   $$
   The unit vector $$ \mathbf{u} $$ in the direction of $$ \mathbf{w} $$ is:
   $$
   \mathbf{u} = \frac{\mathbf{w}}{\|\mathbf{w}\|} = \frac{\mathbf{i} + \mathbf{j}}{\sqrt{2}} = \frac{1}{\sqrt{2}} \mathbf{i} + \frac{1}{\sqrt{2}} \mathbf{j}
   $$

2. **Project $$ \mathbf{v} $$ onto $$ \mathbf{w} $$** to find $$ \mathbf{v}_{1} $$:
   The projection formula is given by:
   $$
   \text{proj}_{\mathbf{w}} \mathbf{v} = \left( \frac{\mathbf{v} \cdot \mathbf{w}}{\mathbf{w} \cdot \mathbf{w}} \right) \mathbf{w}
   $$
   First, we calculate $$ \mathbf{v} \cdot \mathbf{w} $$:
   $$
   \mathbf{v} \cdot \mathbf{w} = (4 \mathbf{i} - 6 \mathbf{j}) \cdot (\mathbf{i} + \mathbf{j}) = 4 \cdot 1 + (-6) \cdot 1 = 4 - 6 = -2
   $$
   Next, we calculate $$ \mathbf{w} \cdot \mathbf{w} $$:
   $$
   \mathbf{w} \cdot \mathbf{w} = (\mathbf{i} + \mathbf{j}) \cdot (\mathbf{i} + \mathbf{j}) = 1 + 1 = 2
   $$
   Now we can find the projection:
   $$
   \text{proj}_{\mathbf{w}} \mathbf{v} = \left( \frac{-2}{2} \right) \mathbf{w} = -1 \cdot (\mathbf{i} + \mathbf{j}) = -\mathbf{i} - \mathbf{j}
   $$
   Thus, $$ \mathbf{v}_{1} = -\mathbf{i} - \mathbf{j} $$.

3. **Find $$ \mathbf{v}_{2} $$**:
   Since $$ \mathbf{v} = \mathbf{v}_{1} + \mathbf{v}_{2} $$, we can find $$ \mathbf{v}_{2} $$ by rearranging:
   $$
   \mathbf{v}_{2} = \mathbf{v} - \mathbf{v}_{1} = (4 \mathbf{i} - 6 \mathbf{j}) - (-\mathbf{i} - \mathbf{j}) = 4 \mathbf{i} - 6 \mathbf{j} + \mathbf{i} + \mathbf{j}
   $$
   Simplifying this gives:
   $$
   \mathbf{v}_{2} = (4 + 1) \mathbf{i} + (-6 + 1) \mathbf{j} = 5 \mathbf{i} - 5 \mathbf{j}
   $$

4. **Final Result**:
   The decomposition of $$ \mathbf{v} $$ is:
   $$
   \mathbf{v}_{1} = -\mathbf{i} - \mathbf{j}, \quad \mathbf{v}_{2} = 5 \mathbf{i} - 5 \mathbf{j}
   $$

Thus, the final answer is:
$$
\mathbf{v}_{1} = -\mathbf{i} - \mathbf{j}, \quad \mathbf{v}_{2} = 5 \mathbf{i} - 5 \mathbf{j}
$$
$$ \mathbf{v}_{1} = -\mathbf{i} - \mathbf{j}, \quad \mathbf{v}_{2} = 5 \mathbf{i} - 5 \mathbf{j} $$

Decompose \( \mathbf{v} \) into two vectors \( \mathbf{v}_{1} \) and \( \mathbf{v}_{2} \), where \( \mathbf{v}_{1} \) is parallel to \( \mathbf{w} \) and \( \mathbf{v}_{2} \) is orthogonal to \( \mathbf{w} \). \( \mathbf{v}=4 \mathbf{i}-6 \mathbf{j} \) and \( \mathbf{w}=\mathbf{i}+\mathbf{j} \)

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