Key Concepts and Formulas

1. Vector Triple Product Identity: The identity for the vector triple product is $\vec{A} \times (\vec{B} \times \vec{C}) = (\vec{A} \cdot \vec{C})\vec{B} - (\vec{A} \cdot \vec{B})\vec{C}$. A crucial special case when $\vec{A} = \vec{C}$ is $\vec{A} \times (\vec{B} \times \vec{A}) = |\vec{A}|^2 \vec{B} - (\vec{A} \cdot \vec{B})\vec{A}$.

2. Properties of Mutually Perpendicular Vectors of Equal Magnitude: Given three vectors $\vec{a}, \vec{b}, \vec{c}$ that are mutually perpendicular and have the same magnitude, say $|\vec{a}| = |\vec{b}| = |\vec{c}| = k$. This implies:

   • $\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{c} = \vec{c} \cdot \vec{a} = 0$.
   • $|\vec{a}|^2 = \vec{a} \cdot \vec{a} = k^2$, and similarly $|\vec{b}|^2 = k^2$, $|\vec{c}|^2 = k^2$.
   • These vectors form an orthogonal basis for 3D space. Any vector $\vec{r}$ can be uniquely represented as $\vec{r} = x\vec{a} + y\vec{b} + z\vec{c}$, where $x = \frac{\vec{a} \cdot \vec{r}}{|\vec{a}|^2}$, $y = \frac{\vec{b} \cdot \vec{r}}{|\vec{b}|^2}$, and $z = \frac{\vec{c} \cdot \vec{r}}{|\vec{c}|^2}$.

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Step-by-Step Solution

We are given the equation: $\overrightarrow a \times \{ (\overrightarrow r - \overrightarrow b ) \times \overrightarrow a \} + \overrightarrow b \times \{ (\overrightarrow r - \overrightarrow c ) \times \overrightarrow b \} + \overrightarrow c \times \{ (\overrightarrow r - \overrightarrow a ) \times \overrightarrow c \} = \overrightarrow 0$

Let's simplify each term individually using the vector triple product identity $\vec{A} \times (\vec{B} \times \vec{A}) = |\vec{A}|^2 \vec{B} - (\vec{A} \cdot \vec{B})\vec{A}$. Let $k^2 = |\vec{a}|^2 = |\vec{b}|^2 = |\vec{c}|^2$.

Step 1: Simplify the first term. The first term is $\vec{a} \times \{ (\vec{r} - \vec{b} ) \times \vec{a} \}$. Applying the identity with $\vec{A} = \vec{a}$ and $\vec{B} = (\vec{r} - \vec{b})$: $\vec{a} \times \{ (\vec{r} - \vec{b} ) \times \vec{a} \} = |\vec{a}|^2 (\vec{r} - \vec{b}) - (\vec{a} \cdot (\vec{r} - \vec{b}))\vec{a}$ Using the distributive property of the dot product and the fact that $\vec{a} \cdot \vec{b} = 0$: $= k^2 (\vec{r} - \vec{b}) - (\vec{a} \cdot \vec{r} - \vec{a} \cdot \vec{b})\vec{a}$ $= k^2 \vec{r} - k^2 \vec{b} - (\vec{a} \cdot \vec{r} - 0)\vec{a}$ $= k^2 \vec{r} - k^2 \vec{b} - (\vec{a} \cdot \vec{r})\vec{a}$

Step 2: Simplify the second term. The second term is $\vec{b} \times \{ (\vec{r} - \vec{c} ) \times \vec{b} \}$. Applying the identity with $\vec{A} = \vec{b}$ and $\vec{B} = (\vec{r} - \vec{c})$: $\vec{b} \times \{ (\vec{r} - \vec{c} ) \times \vec{b} \} = |\vec{b}|^2 (\vec{r} - \vec{c}) - (\vec{b} \cdot (\vec{r} - \vec{c}))\vec{b}$ Using $\vec{b} \cdot \vec{c} = 0$: $= k^2 (\vec{r} - \vec{c}) - (\vec{b} \cdot \vec{r} - \vec{b} \cdot \vec{c})\vec{b}$ $= k^2 \vec{r} - k^2 \vec{c} - (\vec{b} \cdot \vec{r})\vec{b}$

Step 3: Simplify the third term. The third term is $\vec{c} \times \{ (\vec{r} - \vec{a} ) \times \vec{c} \}$. Applying the identity with $\vec{A} = \vec{c}$ and $\vec{B} = (\vec{r} - \vec{a})$: $\vec{c} \times \{ (\vec{r} - \vec{a} ) \times \vec{c} \} = |\vec{c}|^2 (\vec{r} - \vec{a}) - (\vec{c} \cdot (\vec{r} - \vec{a}))\vec{c}$ Using $\vec{c} \cdot \vec{a} = 0$: $= k^2 (\vec{r} - \vec{a}) - (\vec{c} \cdot \vec{r} - \vec{c} \cdot \vec{a})\vec{c}$ $= k^2 \vec{r} - k^2 \vec{a} - (\vec{c} \cdot \vec{r})\vec{c}$

Step 4: Sum the simplified terms and substitute into the original equation. Summing the three simplified terms: $(k^2 \vec{r} - k^2 \vec{b} - (\vec{a} \cdot \vec{r})\vec{a}) + (k^2 \vec{r} - k^2 \vec{c} - (\vec{b} \cdot \vec{r})\vec{b}) + (k^2 \vec{r} - k^2 \vec{a} - (\vec{c} \cdot \vec{r})\vec{c}) = \vec{0}$ Combine terms: $3k^2 \vec{r} - k^2 (\vec{a} + \vec{b} + \vec{c}) - [(\vec{a} \cdot \vec{r})\vec{a} + (\vec{b} \cdot \vec{r})\vec{b} + (\vec{c} \cdot \vec{r})\vec{c}] = \vec{0}$

Step 5: Utilize the orthogonal basis property. Since $\vec{a}, \vec{b}, \vec{c}$ form an orthogonal basis, we can write $\vec{r} = x\vec{a} + y\vec{b} + z\vec{c}$. The dot products are: $\vec{a} \cdot \vec{r} = x|\vec{a}|^2 = xk^2$ $\vec{b} \cdot \vec{r} = y|\vec{b}|^2 = yk^2$ $\vec{c} \cdot \vec{r} = z|\vec{c}|^2 = zk^2$ Substitute these into the bracketed term: $(\vec{a} \cdot \vec{r})\vec{a} + (\vec{b} \cdot \vec{r})\vec{b} + (\vec{c} \cdot \vec{r})\vec{c} = (xk^2)\vec{a} + (yk^2)\vec{b} + (zk^2)\vec{c}$ $= k^2 (x\vec{a} + y\vec{b} + z\vec{c})$ Since $x\vec{a} + y\vec{b} + z\vec{c} = \vec{r}$, the bracketed term simplifies to $k^2 \vec{r}$.

Step 6: Substitute back and solve for $\vec{r}$. Substitute $k^2 \vec{r}$ back into the equation from Step 4: $3k^2 \vec{r} - k^2 (\vec{a} + \vec{b} + \vec{c}) - k^2 \vec{r} = \vec{0}$ Combine the $\vec{r}$ terms: $(3k^2 - k^2) \vec{r} - k^2 (\vec{a} + \vec{b} + \vec{c}) = \vec{0}$ $2k^2 \vec{r} - k^2 (\vec{a} + \vec{b} + \vec{c}) = \vec{0}$ Since $k \ne 0$ (as the vectors have magnitude), $k^2 \ne 0$. We can divide the entire equation by $k^2$: $2\vec{r} - (\vec{a} + \vec{b} + \vec{c}) = \vec{0}$ Rearrange to solve for $\vec{r}$: $2\vec{r} = \vec{a} + \vec{b} + \vec{c}$ $\vec{r} = \frac{1}{2}(\vec{a} + \vec{b} + \vec{c})$

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Common Mistakes & Tips

• Vector Triple Product Application: Ensure you correctly identify the vectors in the $\vec{A} \times (\vec{B} \times \vec{C})$ form. Using the specific form $\vec{A} \times (\vec{B} \times \vec{A})$ is often more direct for this type of problem.
• Orthogonality Conditions: Always remember that $\vec{a} \cdot \vec{b} = 0$ and $|\vec{a}|^2 = k^2$ for mutually perpendicular vectors of equal magnitude. These are critical for simplification.
• Algebraic Errors: Keep track of signs and coefficients carefully when expanding and combining terms. The substitution of $\vec{r} = x\vec{a} + y\vec{b} + z\vec{c}$ is a powerful tool to resolve the dot product terms.

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Summary

The problem requires a careful application of the vector triple product identity and the properties of mutually perpendicular vectors of equal magnitude. By simplifying each term of the given vector equation using the identity and the orthogonality conditions, we arrive at a simplified equation involving $\vec{r}$ and the basis vectors. Expressing $\vec{r}$ in terms of the basis vectors allows us to resolve the terms involving dot products. This leads to a solvable linear equation for $\vec{r}$, yielding $\vec{r} = \frac{1}{2}(\vec{a} + \vec{b} + \vec{c})$.

The final answer is $\boxed{\text{(C)} \frac{1}{2}(\overrightarrow a + \overrightarrow b + \overrightarrow c)}$.