Question:

Step 1: Simplify the first term. The first term is $\hat{i} \times [(\vec{a} - \hat{i})\hat{j} \times \hat{i}]$. Given $\vec{a} = x\hat{i} + y\hat{j} + z\hat{k}$, we first evaluate the scalar product $(\vec{a} - \hat{i})\hat{j}$: $$(\vec{a} - \hat{i})\hat{j} = (x\hat{i} + y\hat{j} + z\hat{k} - \hat{i})\hat{j} = ((x-1)\hat{i} + y\hat{j} + z\hat{k})\hat{j} = y$$ Next, we evaluate the vector product $\hat{j} \times \hat{i}$: $$\hat{j} \times \hat{i} = -\hat{k}$$ Substituting these results into the first term yields: $$\hat{i} \times [y(-\hat{k})] = -y(\hat{i} \times \hat{k}) = -y(-\hat{j}) = y\hat{j}$$ Step 2: Simplify the second term. The second term is $\hat{j} \times [(\vec{a} - \hat{k})\hat{j}]$. First, evaluate the scalar product $(\vec{a} - \hat{k})\hat{j}$: $$(\vec{a} - \hat{k})\hat{j} = (x\hat{i} + y\hat{j} + z\hat{k} - \hat{k})\hat{j} = (x\hat{i} + y\hat{j} + (z-1)\hat{k})\hat{j} = y$$ Substituting this result into the second term yields: $$\hat{j} \times [y\hat{j}] = y(\hat{j} \times \hat{j}) = y(\vec{0}) = \vec{0}$$ Step 3: Simplify the third term. The third term is $\hat{k} \times [(\vec{a} - \hat{i})\hat{k}]$. First, evaluate the scalar product $(\vec{a} - \hat{i})\hat{k}$: $$(\vec{a} - \hat{i})\hat{k} = (x\hat{i} + y\hat{j} + z\hat{k} - \hat{i})\hat{k} = ((x-1)\hat{i} + y\hat{j} + z\hat{k})\hat{k} = z$$ Substituting this result into the third term yields: $$\hat{k} \times [z\hat{k}] = z(\hat{k} \times \hat{k}) = z(\vec{0}) = \vec{0}$$ Step 4: Combine all terms. Substitute the simplified terms back into the original equation: $$y\hat{j} + \vec{0} + \vec{0} = \vec{0}$$ $$y\hat{j} = \vec{0}$$ Since $\hat{j}$ is a non-zero vector, its scalar coefficient must be zero. Therefore, $y=0$.