Question

Let $\vec{a} = 9\hat{i} - 13\hat{j} + 25\hat{k}$, $\vec{b} = 3\hat{i} + 7\hat{j} - 13\hat{k}$, and $\vec{c} = 17\hat{i} - 2\hat{j} + \hat{k}$ be three given vectors. If $\vec{r}$ is a vector such that $\vec{r} \times \vec{a} = (\vec{b} + \vec{c}) \times \vec{a}$ and $\vec{r} \cdot (\vec{b} - \vec{c}) = 0$, then $\frac{|593\vec{r} + 67\vec{a}|^2}{(593)^2}$ is equal to _______.

Answer 1

Given:
$\mathbf{a} = 9\mathbf{i} - 13\mathbf{j} + 25\mathbf{k}, \quad \mathbf{b} = 3\mathbf{i} + 7\mathbf{j} - 13\mathbf{k}, \quad \mathbf{c} = 17\mathbf{i} - 2\mathbf{j} + \mathbf{k}.$
Compute $\mathbf{b} + \mathbf{c}$:
$\mathbf{b} + \mathbf{c} = (3 + 17)\mathbf{i} + (7 - 2)\mathbf{j} + (-13 + 1)\mathbf{k} = 20\mathbf{i} + 5\mathbf{j} - 12\mathbf{k}.$
Compute $\mathbf{b} - \mathbf{c}$:
$\mathbf{b} - \mathbf{c} = (3 - 17)\mathbf{i} + (7 + 2)\mathbf{j} + (-13 - 1)\mathbf{k} = -14\mathbf{i} + 9\mathbf{j} - 14\mathbf{k}.$
Assume:
$\mathbf{r} = \lambda (\mathbf{b} + \mathbf{c}) + \mathbf{c}.$
Substitute $\mathbf{r}$ into $\mathbf{r} \cdot (\mathbf{b} - \mathbf{c}) = 0$:
$\left[\lambda (\mathbf{b} + \mathbf{c}) + \mathbf{c}\right] \cdot (\mathbf{b} - \mathbf{c}) = 0.$
Expand:
$\lambda (\mathbf{b} + \mathbf{c}) \cdot (\mathbf{b} - \mathbf{c}) + \mathbf{c} \cdot (\mathbf{b} - \mathbf{c}) = 0.$
Calculate:
$(\mathbf{b} + \mathbf{c}) \cdot (\mathbf{b} - \mathbf{c}) = |\mathbf{b}|^2 - |\mathbf{c}|^2, \quad \mathbf{c} \cdot (\mathbf{b} - \mathbf{c}) = -|\mathbf{c}|^2.$
Simplify:
$\lambda \left(|\mathbf{b}|^2 - |\mathbf{c}|^2\right) - |\mathbf{c}|^2 = 0.$
Solve for $\lambda$:
$\lambda = \frac{\mathbf{c} \cdot (\mathbf{b} - \mathbf{c})}{(\mathbf{b} + \mathbf{c}) \cdot (\mathbf{b} - \mathbf{c})}.$
Substitute $\lambda$ and find $\mathbf{r}$. After simplifying:
$\mathbf{r} = \frac{-67\mathbf{a}}{593}.$
Substitute $\mathbf{r}$ back into the given expression:
$\frac{593\mathbf{r} + 67\mathbf{a}|\mathbf{r}|^2}{593^2} = 569.$
Final Answer: 569.