1. Key Concepts and Formulas

• Vector Projection: The projection of vector $\overrightarrow{b}$ onto vector $\overrightarrow{a}$ is given by $\text{proj}_{\overrightarrow{a}}\overrightarrow{b} = \frac{\overrightarrow{b} \cdot \overrightarrow{a}}{||\overrightarrow{a}||^2}\overrightarrow{a}$. This component is parallel to $\overrightarrow{a}$.
• Vector Decomposition: A vector $\overrightarrow{b}$ can be decomposed into a component parallel to $\overrightarrow{a}$ and a component perpendicular to $\overrightarrow{a}$. If $\overrightarrow{b} = \overrightarrow{b}_{||} + \overrightarrow{b}_{\perp}$, where $\overrightarrow{b}_{||}$ is parallel to $\overrightarrow{a}$ and $\overrightarrow{b}_{\perp}$ is perpendicular to $\overrightarrow{a}$, then $\overrightarrow{b}_{||} = \text{proj}_{\overrightarrow{a}}\overrightarrow{b}$ and $\overrightarrow{b}_{\perp} = \overrightarrow{b} - \overrightarrow{b}_{||}$.
• Vector Cross Product: The cross product of two vectors $\overrightarrow{u} = u_1\widehat i + u_2\widehat j + u_3\widehat k$ and $\overrightarrow{v} = v_1\widehat i + v_2\widehat j + v_3\widehat k$ is given by the determinant: $\overrightarrow{u} \times \overrightarrow{v} = \begin{vmatrix} \widehat i & \widehat j & \widehat k \\ u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \end{vmatrix}$

2. Step-by-Step Solution

Step 1: Identify the components of $\overrightarrow{\beta}$ We are given $\overrightarrow \beta = {\overrightarrow \beta _1} - \overrightarrow {{\beta _2}}$. We are also given that ${\overrightarrow \beta _1}$ is parallel to $\overrightarrow \alpha$ and $\overrightarrow {{\beta _2}}$ is perpendicular to $\overrightarrow \alpha$. This means that ${\overrightarrow \beta _1}$ is the component of $\overrightarrow \beta$ that lies in the direction of $\overrightarrow \alpha$, and $\overrightarrow {{\beta _2}}$ is related to the component of $\overrightarrow \beta$ perpendicular to $\overrightarrow \alpha$.

Step 2: Calculate ${\overrightarrow \beta _1}$ Since ${\overrightarrow \beta _1}$ is parallel to $\overrightarrow \alpha$ and is a component of $\overrightarrow \beta$, it must be the vector projection of $\overrightarrow \beta$ onto $\overrightarrow \alpha$. We are given $\overrightarrow \alpha = 3\widehat i + \widehat j$ and $\overrightarrow \beta = 2\widehat i - \widehat j + 3 \widehat k$.

First, calculate the dot product $\overrightarrow \beta \cdot \overrightarrow \alpha$: $\overrightarrow \beta \cdot \overrightarrow \alpha = (2)(3) + (-1)(1) + (3)(0) = 6 - 1 + 0 = 5$

Next, calculate the square of the magnitude of $\overrightarrow \alpha$: $||\overrightarrow \alpha||^2 = (3)^2 + (1)^2 + (0)^2 = 9 + 1 = 10$

Now, use the vector projection formula to find ${\overrightarrow \beta _1}$: ${\overrightarrow \beta _1} = \frac{\overrightarrow \beta \cdot \overrightarrow \alpha}{||\overrightarrow \alpha||^2}\overrightarrow \alpha = \frac{5}{10}(3\widehat i + \widehat j) = \frac{1}{2}(3\widehat i + \widehat j) = \frac{3}{2}\widehat i + \frac{1}{2}\widehat j$

Step 3: Calculate $\overrightarrow {{\beta _2}}$ From the given decomposition $\overrightarrow \beta = {\overrightarrow \beta _1} - \overrightarrow {{\beta _2}}$, we can rearrange to solve for $\overrightarrow {{\beta _2}}$: $\overrightarrow {{\beta _2}} = {\overrightarrow \beta _1} - \overrightarrow \beta$

Substitute the calculated value of ${\overrightarrow \beta _1}$ and the given $\overrightarrow \beta$: $\overrightarrow {{\beta _2}} = \left(\frac{3}{2}\widehat i + \frac{1}{2}\widehat j\right) - (2\widehat i - \widehat j + 3 \widehat k)$

Perform the vector subtraction: $\overrightarrow {{\beta _2}} = \left(\frac{3}{2} - 2\right)\widehat i + \left(\frac{1}{2} - (-1)\right)\widehat j + (0 - 3)\widehat k$ $\overrightarrow {{\beta _2}} = \left(\frac{3 - 4}{2}\right)\widehat i + \left(\frac{1 + 2}{2}\right)\widehat j - 3\widehat k$ $\overrightarrow {{\beta _2}} = -\frac{1}{2}\widehat i + \frac{3}{2}\widehat j - 3\widehat k$ We can verify that $\overrightarrow {{\beta _2}}$ is indeed perpendicular to $\overrightarrow \alpha$: $\overrightarrow {{\beta _2}} \cdot \overrightarrow \alpha = (-\frac{1}{2})(3) + (\frac{3}{2})(1) + (-3)(0) = -\frac{3}{2} + \frac{3}{2} + 0 = 0$ This confirms our calculation for $\overrightarrow {{\beta _2}}$ is consistent with the problem statement.

Step 4: Calculate the cross product ${\overrightarrow \beta _1} \times \overrightarrow {{\beta _2}}$ Now we compute the cross product of ${\overrightarrow \beta _1} = \frac{3}{2}\widehat i + \frac{1}{2}\widehat j + 0\widehat k$ and $\overrightarrow {{\beta _2}} = -\frac{1}{2}\widehat i + \frac{3}{2}\widehat j - 3\widehat k$.

Using the determinant formula for the cross product: ${\overrightarrow \beta _1} \times \overrightarrow {{\beta _2}} = \begin{vmatrix} \widehat i & \widehat j & \widehat k \\ \frac{3}{2} & \frac{1}{2} & 0 \\ -\frac{1}{2} & \frac{3}{2} & -3 \end{vmatrix}$

Expand the determinant: ${\overrightarrow \beta _1} \times \overrightarrow {{\beta _2}} = \widehat i \left( (\frac{1}{2})(-3) - (0)(\frac{3}{2}) \right) - \widehat j \left( (\frac{3}{2})(-3) - (0)(-\frac{1}{2}) \right) + \widehat k \left( (\frac{3}{2})(\frac{3}{2}) - (\frac{1}{2})(-\frac{1}{2}) \right)$ ${\overrightarrow \beta _1} \times \overrightarrow {{\beta _2}} = \widehat i \left( -\frac{3}{2} - 0 \right) - \widehat j \left( -\frac{9}{2} - 0 \right) + \widehat k \left( \frac{9}{4} - (-\frac{1}{4}) \right)$ ${\overrightarrow \beta _1} \times \overrightarrow {{\beta _2}} = -\frac{3}{2}\widehat i + \frac{9}{2}\widehat j + \widehat k \left( \frac{9}{4} + \frac{1}{4} \right)$ ${\overrightarrow \beta _1} \times \overrightarrow {{\beta _2}} = -\frac{3}{2}\widehat i + \frac{9}{2}\widehat j + \frac{10}{4}\widehat k$ ${\overrightarrow \beta _1} \times \overrightarrow {{\beta _2}} = -\frac{3}{2}\widehat i + \frac{9}{2}\widehat j + \frac{5}{2}\widehat k$

Factor out $\frac{1}{2}$: ${\overrightarrow \beta _1} \times \overrightarrow {{\beta _2}} = \frac{1}{2}(-3\widehat i + 9\widehat j + 5\widehat k)$

3. Common Mistakes & Tips

• Interpreting the Decomposition: The given decomposition is $\overrightarrow \beta = {\overrightarrow \beta _1} - \overrightarrow {{\beta _2}}$. This is crucial. If one mistakenly assumes $\overrightarrow \beta = {\overrightarrow \beta _1} + \overrightarrow {{\beta _2}}$, then $\overrightarrow {{\beta _2}}$ will be the standard perpendicular component (vector rejection), leading to a sign difference in the final answer. Always derive $\overrightarrow {{\beta _2}}$ directly from the provided equation.
• Vector Projection Formula: Ensure the correct formula for vector projection is used, which includes multiplying by the vector $\overrightarrow{a}$ itself, not just its unit vector.
• Cross Product Calculation: Double-check the signs and arithmetic during the determinant expansion for the cross product. A small error here can lead to a completely different result.

4. Summary

The problem requires decomposing $\overrightarrow \beta$ into two components, ${\overrightarrow \beta _1}$ parallel to $\overrightarrow \alpha$ and $\overrightarrow {{\beta _2}}$ which, when subtracted from ${\overrightarrow \beta _1}$, yields $\overrightarrow \beta$. We first calculated ${\overrightarrow \beta _1}$ as the vector projection of $\overrightarrow \beta$ onto $\overrightarrow \alpha$. Then, using the given relation $\overrightarrow \beta = {\overrightarrow \beta _1} - \overrightarrow {{\beta _2}}$, we derived $\overrightarrow {{\beta _2}} = {\overrightarrow \beta _1} - \overrightarrow \beta$. Finally, we computed the cross product ${\overrightarrow \beta _1} \times \overrightarrow {{\beta _2}}$ using the determinant method.

The final answer is $\boxed{\text{(A)}}$.