Key Concepts and Formulas

1. Vector Cross Product Properties:

   • $\vec{X} \times \vec{Y} = \vec{0}$ if and only if $\vec{X}$ and $\vec{Y}$ are parallel or one of them is the zero vector.
   • $(\vec{A} - \vec{B}) \times \vec{C} = \vec{A} \times \vec{C} - \vec{B} \times \vec{C}$ (Distributive property).
   • $\vec{A} \times \vec{B} = -(\vec{B} \times \vec{A})$ (Anti-commutative property).

2. Vector Dot Product Properties:

   • $(\vec{A} + \vec{B}) \cdot \vec{C} = \vec{A} \cdot \vec{C} + \vec{B} \cdot \vec{C}$ (Distributive property).
   • $(\lambda \vec{A}) \cdot \vec{B} = \lambda (\vec{A} \cdot \vec{B})$ (Scalar multiple property).
   • $\vec{A} \cdot \vec{B} = 0$ if and only if $\vec{A}$ and $\vec{B}$ are orthogonal (perpendicular).
   • If $\vec{A} = A_x \hat{i} + A_y \hat{j} + A_z \hat{k}$ and $\vec{B} = B_x \hat{i} + B_y \hat{j} + B_z \hat{k}$, then $\vec{A} \cdot \vec{B} = A_x B_x + A_y B_y + A_z B_z$.

3. Magnitude of a Vector: If $\vec{V} = V_x \hat{i} + V_y \hat{j} + V_z \hat{k}$, then its magnitude is $|\vec{V}| = \sqrt{V_x^2 + V_y^2 + V_z^2}$.

Step-by-Step Solution

Step 1: Simplify the first given condition using the cross product property. We are given the equation $\vec{r} \times \vec{a}=\vec{c} \times \vec{a}$. To simplify this, we can rearrange it by moving all terms to one side: $\vec{r} \times \vec{a} - \vec{c} \times \vec{a} = \vec{0}$ Using the distributive property of the cross product, $(\vec{A} - \vec{B}) \times \vec{C} = \vec{A} \times \vec{C} - \vec{B} \times \vec{C}$, we can write: $(\vec{r} - \vec{c}) \times \vec{a} = \vec{0}$ Reasoning: The cross product of two vectors is the zero vector if and only if the vectors are parallel (or one is the zero vector). Since $\vec{a}=2 \hat{i}-7 \hat{j}+5 \hat{k}$ is a non-zero vector, the vector $(\vec{r} - \vec{c})$ must be parallel to $\vec{a}$. This parallelism implies that $(\vec{r} - \vec{c})$ can be expressed as a scalar multiple of $\vec{a}$: $\vec{r} - \vec{c} = \lambda \vec{a}$ where $\lambda$ is a scalar. Rearranging this equation to express $\vec{r}$ in terms of $\vec{a}$, $\vec{c}$, and $\lambda$: $\vec{r} = \vec{c} + \lambda \vec{a} \quad \text{(Equation 1)}$ This equation provides a general form for vector $\vec{r}$.

Step 2: Use the second given condition to find the scalar $\lambda$. We are given the second condition $\vec{r} \cdot \vec{b} = 0$. This means that $\vec{r}$ is orthogonal to $\vec{b}$. Substitute the expression for $\vec{r}$ from Equation 1 into this dot product equation: $(\vec{c} + \lambda \vec{a}) \cdot \vec{b} = 0$ Using the distributive and scalar multiple properties of the dot product: $\vec{c} \cdot \vec{b} + (\lambda \vec{a}) \cdot \vec{b} = 0$ $\vec{c} \cdot \vec{b} + \lambda (\vec{a} \cdot \vec{b}) = 0$ Now, we need to calculate the dot products $\vec{c} \cdot \vec{b}$ and $\vec{a} \cdot \vec{b}$. Given vectors are: $\vec{a} = 2 \hat{i} - 7 \hat{j} + 5 \hat{k}$ $\vec{b} = \hat{i} + \hat{k}$ $\vec{c} = \hat{i} + 2 \hat{j} - 3 \hat{k}$

Calculate $\vec{c} \cdot \vec{b}$: $\vec{c} \cdot \vec{b} = (1)(\hat{i} + 2 \hat{j} - 3 \hat{k}) \cdot (\hat{i} + 0 \hat{j} + 1 \hat{k})$ $\vec{c} \cdot \vec{b} = (1)(1) + (2)(0) + (-3)(1) = 1 + 0 - 3 = -2$

Calculate $\vec{a} \cdot \vec{b}$: $\vec{a} \cdot \vec{b} = (2 \hat{i} - 7 \hat{j} + 5 \hat{k}) \cdot (\hat{i} + 0 \hat{j} + 1 \hat{k})$ $\vec{a} \cdot \vec{b} = (2)(1) + (-7)(0) + (5)(1) = 2 + 0 + 5 = 7$

Substitute these values back into the equation: $-2 + \lambda (7) = 0$ $7\lambda = 2$ $\lambda = \frac{2}{7}$

Step 3: Find the vector $\vec{r}$ and its magnitude. Now that we have found the value of $\lambda$, we can substitute it back into Equation 1 to find the explicit form of $\vec{r}$: $\vec{r} = \vec{c} + \lambda \vec{a}$ $\vec{r} = (\hat{i} + 2 \hat{j} - 3 \hat{k}) + \frac{2}{7} (2 \hat{i} - 7 \hat{j} + 5 \hat{k})$ Distribute the scalar $\frac{2}{7}$: $\vec{r} = \hat{i} + 2 \hat{j} - 3 \hat{k} + \frac{4}{7} \hat{i} - \frac{14}{7} \hat{j} + \frac{10}{7} \hat{k}$ $\vec{r} = \hat{i} + 2 \hat{j} - 3 \hat{k} + \frac{4}{7} \hat{i} - 2 \hat{j} + \frac{10}{7} \hat{k}$ Group the $\hat{i}$, $\hat{j}$, and $\hat{k}$ components: $\vec{r} = \left(1 + \frac{4}{7}\right) \hat{i} + \left(2 - 2\right) \hat{j} + \left(-3 + \frac{10}{7}\right) \hat{k}$ $\vec{r} = \left(\frac{7+4}{7}\right) \hat{i} + (0) \hat{j} + \left(\frac{-21+10}{7}\right) \hat{k}$ $\vec{r} = \frac{11}{7} \hat{i} + 0 \hat{j} - \frac{11}{7} \hat{k}$ So, $\vec{r} = \frac{11}{7} \hat{i} - \frac{11}{7} \hat{k}$.

Finally, we need to find the magnitude of $\vec{r}$: $|\vec{r}| = \left|\frac{11}{7} \hat{i} - \frac{11}{7} \hat{k}\right|$ $|\vec{r}| = \sqrt{\left(\frac{11}{7}\right)^2 + (0)^2 + \left(-\frac{11}{7}\right)^2}$ $|\vec{r}| = \sqrt{\frac{121}{49} + 0 + \frac{121}{49}}$ $|\vec{r}| = \sqrt{\frac{2 \times 121}{49}}$ $|\vec{r}| = \frac{\sqrt{2 \times 121}}{\sqrt{49}}$ $|\vec{r}| = \frac{11\sqrt{2}}{7}$

Common Mistakes & Tips

• Algebraic Errors: Be meticulous with algebraic manipulations, especially when dealing with fractions and signs in vector operations. Double-check calculations of dot products and vector additions/subtractions.
• Misinterpreting Cross Product: Remember that $\vec{X} \times \vec{Y} = \vec{0}$ implies parallelism, not necessarily equality. This is the key insight to derive $\vec{r} = \vec{c} + \lambda \vec{a}$.
• Order of Operations: Ensure that vector operations (cross product, dot product) are performed correctly and in the right order.

Summary

The problem involves finding a vector $\vec{r}$ that satisfies two conditions: it is parallel to $\vec{c}$ plus a scalar multiple of $\vec{a}$ (derived from the cross product condition), and it is orthogonal to $\vec{b}$ (from the dot product condition). By substituting the general form of $\vec{r}$ into the dot product equation, we solved for the scalar $\lambda$. Once $\lambda$ was determined, we found the explicit vector $\vec{r}$ and then calculated its magnitude. The magnitude of $\vec{r}$ was found to be $\frac{11\sqrt{2}}{7}$.

The final answer is $\boxed{\frac{11 \sqrt{2}}{7}}$. This corresponds to option (D).