For any vector \vec{r} , what is (\vec{r}\cdot\hat{i})(\vec{r}\times\hat{i})+(\vec{r}\cdot\hat{j})(\vec{r}\times\hat{j})+(\vec{r}\cdot\hat{k})(\vec{r}\times\hat{k}) equal to ?

  1. A. \vec{0} ✓
  2. B. \vec{r}
  3. C. 2\vec{r}
  4. D. 3\vec{r}

Correct Answer: A. \vec{0}

Explanation

Let \vec{r} = x\hat{i} + y\hat{j} + z\hat{k}. Then \vec{r}\cdot\hat{i} = x, and \vec{r}\times\hat{i} = (x\hat{i} + y\hat{j} + z\hat{k}) \times \hat{i} = -y\hat{k} + z\hat{j}. Thus, the first term is x(-y\hat{k} + z\hat{j}) = -xy\hat{k} + xz\hat{j}. Similarly, the second term is y(x\hat{k} - z\hat{i}) = xy\hat{k} - yz\hat{i}, and the third term is z(-x\hat{j} + y\hat{i}) = -xz\hat{j} + yz\hat{i}. Adding all three terms together: (-xy\hat{k} + xz\hat{j}) + (xy\hat{k} - yz\hat{i}) + (-xz\hat{j} + yz\hat{i}) = \vec{0}.

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