Question

Let $\vec{a} = \hat{i} + 2\hat{j} - \hat{k}$, $\vec{b} = \hat{i} - \hat{j}$ and $\vec{c} = \hat{i} - \hat{j} - \hat{k}$. If $\vec{r}$ is a vector such that $\vec{r} \times \vec{a} = \vec{c} \times \vec{a}$ and $\vec{r} \cdot \vec{b} = 0$, then $\vec{r} \cdot \vec{a}$ is equal to __________.

Hint:

If $(\vec{u} - \vec{v}) \times \vec{w} = 0$, you can always write $\vec{u} = \vec{v} + \lambda \vec{w}$. This is the most common way to solve vector equation problems.

Answer 1

Correct Answer: 12

Step 1: $\vec{r} \times \vec{a} - \vec{c} \times \vec{a} = 0 \Rightarrow (\vec{r} - \vec{c}) \times \vec{a} = 0$.
Step 2: This means $\vec{r} - \vec{c}$ is parallel to $\vec{a}$, so $\vec{r} = \vec{c} + \lambda \vec{a}$.
Step 3: Use $\vec{r} \cdot \vec{b} = 0 \Rightarrow (\vec{c} + \lambda \vec{a}) \cdot \vec{b} = 0$.
Step 4: $\vec{c} \cdot \vec{b} = (1)(1) + (-1)(-1) + (-1)(0) = 2$.
Step 5: $\vec{a} \cdot \vec{b} = (1)(1) + (2)(-1) + (-1)(0) = -1$.
Step 6: $2 + \lambda(-1) = 0 \Rightarrow \lambda = 2$.
Step 7: $\vec{r} \cdot \vec{a} = (\vec{c} + 2\vec{a}) \cdot \vec{a} = \vec{c} \cdot \vec{a} + 2|\vec{a}|^2$.
Step 8: $\vec{c} \cdot \vec{a} = (1)(1) + (-1)(2) + (-1)(-1) = 1 - 2 + 1 = 0$.
Step 9: $|\vec{a}|^2 = 1^2 + 2^2 + (-1)^2 = 6$.
Step 10: $\vec{r} \cdot \vec{a} = 0 + 2(6) = 12$.