The vector $$(\hat{i} \times \vec{a} \cdot \vec{b})\hat{i} + (\hat{j} \times \vec{a} \cdot \vec{b})\hat{j} + (\hat{k} \times \vec{a} \cdot \vec{b})\hat{k}$$ is equal to:

The given vector expression is $$ (\hat{i} \times \vec{a} \cdot \vec{b})\hat{i} + (\hat{j} \times \vec{a} \cdot \vec{b})\hat{j} + (\hat{k} \times \vec{a} \cdot \vec{b})\hat{k} $$. Each term involves a cross product followed by a dot product, which is a scalar triple product. Specifically, $$ \hat{i} \times \vec{a} \cdot \vec{b} $$ means $$ (\hat{i} \times \vec{a}) \cdot \vec{b} $$, and by the properties of the scalar triple product, this is equal to $$ \hat{i} \cdot (\vec{a} \times \vec{b}) $$. Similarly, $$ \hat{j} \times \vec{a} \cdot \vec{b} = \hat{j} \cdot (\vec{a} \times \vec{b}) $$ and $$ \hat{k} \times \vec{a} \cdot \vec{b} = \hat{k} \cdot (\vec{a} \times \vec{b}) $$.

Let $$ \vec{c} = \vec{a} \times \vec{b} $$. Then the expression becomes:

$$ (\hat{i} \cdot \vec{c}) \hat{i} + (\hat{j} \cdot \vec{c}) \hat{j} + (\hat{k} \cdot \vec{c}) \hat{k} $$

This is the standard way to express the vector $$ \vec{c} $$ in terms of its components along the Cartesian unit vectors. The component of $$ \vec{c} $$ along $$ \hat{i} $$ is $$ \hat{i} \cdot \vec{c} $$, along $$ \hat{j} $$ is $$ \hat{j} \cdot \vec{c} $$, and along $$ \hat{k} $$ is $$ \hat{k} \cdot \vec{c} $$. Therefore, the entire expression simplifies to $$ \vec{c} $$, which is $$ \vec{a} \times \vec{b} $$.

Now, comparing with the options:

• Option A: $$ \vec{b} \times \vec{a} $$ is the negative of $$ \vec{a} \times \vec{b} $$, so it is not equal.
• Option B: $$ \vec{a} $$ is not necessarily equal to $$ \vec{a} \times \vec{b} $$.
• Option C: $$ \vec{a} \times \vec{b} $$ matches our result.
• Option D: $$ \vec{b} $$ is not necessarily equal to $$ \vec{a} \times \vec{b} $$.

Hence, the correct answer is Option C.