Key Concepts and Formulas

1. Vector Cross Product Property: If $\vec{X} \times \vec{Y} = \vec{0}$ and $\vec{Y} \neq \vec{0}$, then $\vec{X}$ is parallel to $\vec{Y}$, meaning $\vec{X} = \lambda \vec{Y}$ for some scalar $\lambda$. This property is crucial for simplifying vector equations involving cross products.
2. Vector Dot Product Properties: The dot product is distributive ($\vec{A} \cdot (\vec{B} + \vec{C}) = \vec{A} \cdot \vec{B} + \vec{A} \cdot \vec{C}$) and can be calculated using components: $\vec{A} \cdot \vec{B} = A_x B_x + A_y B_y + A_z B_z$.
3. Magnitude of a Vector: For $\vec{v} = v_x \hat{i} + v_y \hat{j} + v_z \hat{k}$, $|\vec{v}|^2 = v_x^2 + v_y^2 + v_z^2$.

Step-by-Step Solution

Step 1: Simplify the cross product condition to express $\vec{r}$.

We are given the condition $\vec{r} \times \vec{b}=\vec{c} \times \vec{b}$. To use the cross product property, we rearrange the equation by moving all terms to one side: $\vec{r} \times \vec{b} - \vec{c} \times \vec{b} = \vec{0}$ Using the distributive property of the cross product, $(\vec{X} - \vec{Y}) \times \vec{Z} = \vec{X} \times \vec{Z} - \vec{Y} \times \vec{Z}$, we can rewrite the equation as: $(\vec{r} - \vec{c}) \times \vec{b} = \vec{0}$ This equation implies that the vector $(\vec{r} - \vec{c})$ is parallel to vector $\vec{b}$. Since $\vec{b} = \hat{i}-\hat{j}+2 \hat{k}$ is a non-zero vector, we can write: $\vec{r} - \vec{c} = \lambda \vec{b}$ where $\lambda$ is a scalar. Rearranging this equation gives us an expression for $\vec{r}$: $\vec{r} = \vec{c} + \lambda \vec{b} \quad \text{(Equation 1)}$ This step is crucial as it expresses the unknown vector $\vec{r}$ in terms of known vectors and a single scalar unknown, $\lambda$.

Step 2: Use the dot product condition to find the scalar $\lambda$.

The second condition given is $\vec{r} \cdot \vec{a}=0$. Substitute the expression for $\vec{r}$ from Equation 1 into this condition: $(\vec{c} + \lambda \vec{b}) \cdot \vec{a} = 0$ Using the distributive property of the dot product, we expand this equation: $\vec{c} \cdot \vec{a} + \lambda (\vec{b} \cdot \vec{a}) = 0$ Now, we calculate the dot products $\vec{c} \cdot \vec{a}$ and $\vec{b} \cdot \vec{a}$. Given $\vec{a}=\hat{i}+2 \hat{j}+3 \hat{k}$, $\vec{b}=\hat{i}-\hat{j}+2 \hat{k}$, and $\vec{c}=5 \hat{i}-3 \hat{j}+3 \hat{k}$:

Calculate $\vec{c} \cdot \vec{a}$: $\vec{c} \cdot \vec{a} = (5)(1) + (-3)(2) + (3)(3) = 5 - 6 + 9 = 8$ Calculate $\vec{b} \cdot \vec{a}$: $\vec{b} \cdot \vec{a} = (1)(1) + (-1)(2) + (2)(3) = 1 - 2 + 6 = 5$ Substitute these values back into the scalar equation: $8 + \lambda (5) = 0$ $5\lambda = -8$ $\lambda = -\frac{8}{5}$ This step successfully determines the scalar value $\lambda$ that satisfies both given conditions for $\vec{r}$.

Step 3: Determine the vector $\vec{r}$.

Now that we have $\lambda = -\frac{8}{5}$, we substitute it back into Equation 1 to find the components of $\vec{r}$: $\vec{r} = \vec{c} + \lambda \vec{b}$ $\vec{r} = (5 \hat{i}-3 \hat{j}+3 \hat{k}) + \left(-\frac{8}{5}\right) (\hat{i}-\hat{j}+2 \hat{k})$ Distribute the scalar $-\frac{8}{5}$ to the components of $\vec{b}$: $\vec{r} = (5 \hat{i}-3 \hat{j}+3 \hat{k}) - \frac{8}{5}\hat{i} + \frac{8}{5}\hat{j} - \frac{16}{5}\hat{k}$ Combine the coefficients of $\hat{i}$, $\hat{j}$, and $\hat{k}$: $\vec{r} = \left(5 - \frac{8}{5}\right)\hat{i} + \left(-3 + \frac{8}{5}\right)\hat{j} + \left(3 - \frac{16}{5}\right)\hat{k}$ $\vec{r} = \left(\frac{25-8}{5}\right)\hat{i} + \left(\frac{-15+8}{5}\right)\hat{j} + \left(\frac{15-16}{5}\right)\hat{k}$ $\vec{r} = \frac{17}{5}\hat{i} - \frac{7}{5}\hat{j} - \frac{1}{5}\hat{k}$ So, $\vec{r} = \frac{1}{5}(17\hat{i} - 7\hat{j} - \hat{k})$.

Step 4: Calculate $25|\vec{r}|^{2}$.

First, we find the magnitude squared of $\vec{r}$: $|\vec{r}|^2 = \left(\frac{17}{5}\right)^2 + \left(-\frac{7}{5}\right)^2 + \left(-\frac{1}{5}\right)^2$ $|\vec{r}|^2 = \frac{289}{25} + \frac{49}{25} + \frac{1}{25}$ Combine the terms: $|\vec{r}|^2 = \frac{289 + 49 + 1}{25} = \frac{339}{25}$ Finally, we calculate $25|\vec{r}|^{2}$: $25|\vec{r}|^{2} = 25 \times \left(\frac{339}{25}\right) = 339$

Common Mistakes & Tips

• Algebraic Precision: Double-check all arithmetic calculations, especially when dealing with fractions and signs, as errors can easily propagate.
• Cross Product Interpretation: Remember that $\vec{X} \times \vec{Y} = \vec{0}$ implies parallelism ($\vec{X} = \lambda \vec{Y}$) for non-zero $\vec{Y}$, not necessarily that $\vec{X}$ or $\vec{Y}$ is zero.
• Systematic Approach: Follow the logical flow of simplifying the cross product first, then using the dot product to solve for the scalar, and finally reconstructing the vector.

Summary

The problem involves finding an unknown vector $\vec{r}$ satisfying conditions related to both cross and dot products. We first simplified the cross product equality $\vec{r} \times \vec{b}=\vec{c} \times \vec{b}$ to deduce that $\vec{r} - \vec{c}$ is parallel to $\vec{b}$, allowing us to express $\vec{r}$ as $\vec{c} + \lambda \vec{b}$. Substituting this into the dot product condition $\vec{r} \cdot \vec{a}=0$ allowed us to solve for the scalar $\lambda$. With $\lambda$ determined, we found the explicit form of $\vec{r}$ and subsequently calculated $25|\vec{r}|^{2}$.

The final answer is $\boxed{339}$.