Key Concepts and Formulas

• Vector Triple Product Identity: The identity for the vector triple product is given by $\vec{A} \times (\vec{B} \times \vec{C}) = (\vec{A} \cdot \vec{C})\vec{B} - (\vec{A} \cdot \vec{B})\vec{C}$. This is a fundamental formula for simplifying expressions involving the cross product of three vectors.
• Properties of Orthonormal Basis Vectors: For the standard basis vectors $\widehat i, \widehat j, \widehat k$:
  • $\widehat i \cdot \widehat i = \widehat j \cdot \widehat j = \widehat k \cdot \widehat k = 1$
  • $\widehat i \cdot \widehat j = \widehat j \cdot \widehat k = \widehat k \cdot \widehat i = 0$
  • $\widehat i \times \widehat j = \widehat k$, $\widehat j \times \widehat k = \widehat i$, $\widehat k \times \widehat i = \widehat j$ (and cyclic permutations).
• Magnitude of a Vector: For a vector $\vec{V} = V_x \widehat i + V_y \widehat j + V_z \widehat k$, its magnitude squared is $|\vec{V}|^2 = V_x^2 + V_y^2 + V_z^2$.

Step-by-Step Solution

We are asked to find the value of the expression $S = {\left| {\widehat i \times \left( {\overrightarrow a \times \widehat i} \right)} \right|^2} + {\left| {\widehat j \times \left( {\overrightarrow a \times \widehat j} \right)} \right|^2} + {\left| {\widehat k \times \left( {\overrightarrow a \times \widehat k} \right)} \right|^2}$. The given vector is $\overrightarrow a = 2\widehat i + \widehat j + 2\widehat k$.

Step 1: Simplify the first term using the Vector Triple Product Identity. Let's focus on the first term: $\widehat i \times (\overrightarrow a \times \widehat i)$. We apply the vector triple product identity $\vec{A} \times (\vec{B} \times \vec{C}) = (\vec{A} \cdot \vec{C})\vec{B} - (\vec{A} \cdot \vec{B})\vec{C}$ with $\vec{A} = \widehat i$, $\vec{B} = \overrightarrow a$, and $\vec{C} = \widehat i$.

$\widehat i \times \left( {\overrightarrow a \times \widehat i} \right) = \left( {\widehat i \cdot \widehat i} \right)\overrightarrow a - \left( {\widehat i \cdot \overrightarrow a } \right)\widehat i$

Now, we evaluate the dot products:

• $\widehat i \cdot \widehat i = 1$ (as it's the dot product of a unit vector with itself).
• $\overrightarrow a = 2\widehat i + \widehat j + 2\widehat k$. So, $\widehat i \cdot \overrightarrow a = \widehat i \cdot (2\widehat i + \widehat j + 2\widehat k) = 2(\widehat i \cdot \widehat i) + 1(\widehat i \cdot \widehat j) + 2(\widehat i \cdot \widehat k)$. Using the properties of orthonormal vectors, $\widehat i \cdot \widehat j = 0$ and $\widehat i \cdot \widehat k = 0$. Thus, $\widehat i \cdot \overrightarrow a = 2(1) + 0 + 0 = 2$.

Substituting these values back into the identity: $\widehat i \times \left( {\overrightarrow a \times \widehat i} \right) = (1)\overrightarrow a - (2)\widehat i = \overrightarrow a - 2\widehat i$ Now, substitute the components of $\overrightarrow a$: $\overrightarrow a - 2\widehat i = (2\widehat i + \widehat j + 2\widehat k) - 2\widehat i = \widehat j + 2\widehat k$ The square of the magnitude of this vector is: ${\left| {\widehat j + 2\widehat k} \right|^2} = (0)^2 + (1)^2 + (2)^2 = 1 + 4 = 5$

Step 2: Simplify the second term using symmetry and the Vector Triple Product Identity. The second term is ${\left| {\widehat j \times \left( {\overrightarrow a \times \widehat j} \right)} \right|^2}$. Applying the vector triple product identity with $\vec{A} = \widehat j$, $\vec{B} = \overrightarrow a$, and $\vec{C} = \widehat j$: $\widehat j \times \left( {\overrightarrow a \times \widehat j} \right) = \left( {\widehat j \cdot \widehat j} \right)\overrightarrow a - \left( {\widehat j \cdot \overrightarrow a } \right)\widehat j$ We know $\widehat j \cdot \widehat j = 1$. And $\widehat j \cdot \overrightarrow a = \widehat j \cdot (2\widehat i + \widehat j + 2\widehat k) = 2(\widehat j \cdot \widehat i) + 1(\widehat j \cdot \widehat j) + 2(\widehat j \cdot \widehat k) = 0 + 1 + 0 = 1$.

Substituting these values: $\widehat j \times \left( {\overrightarrow a \times \widehat j} \right) = (1)\overrightarrow a - (1)\widehat j = \overrightarrow a - \widehat j$ Substitute the components of $\overrightarrow a$: $\overrightarrow a - \widehat j = (2\widehat i + \widehat j + 2\widehat k) - \widehat j = 2\widehat i + 2\widehat k$ The square of the magnitude of this vector is: ${\left| {2\widehat i + 2\widehat k} \right|^2} = (2)^2 + (0)^2 + (2)^2 = 4 + 4 = 8$

Step 3: Simplify the third term using symmetry and the Vector Triple Product Identity. The third term is ${\left| {\widehat k \times \left( {\overrightarrow a \times \widehat k} \right)} \right|^2}$. Applying the vector triple product identity with $\vec{A} = \widehat k$, $\vec{B} = \overrightarrow a$, and $\vec{C} = \widehat k$: $\widehat k \times \left( {\overrightarrow a \times \widehat k} \right) = \left( {\widehat k \cdot \widehat k} \right)\overrightarrow a - \left( {\widehat k \cdot \overrightarrow a } \right)\widehat k$ We know $\widehat k \cdot \widehat k = 1$. And $\widehat k \cdot \overrightarrow a = \widehat k \cdot (2\widehat i + \widehat j + 2\widehat k) = 2(\widehat k \cdot \widehat i) + 1(\widehat k \cdot \widehat j) + 2(\widehat k \cdot \widehat k) = 0 + 0 + 2(1) = 2$.

Substituting these values: $\widehat k \times \left( {\overrightarrow a \times \widehat k} \right) = (1)\overrightarrow a - (2)\widehat k = \overrightarrow a - 2\widehat k$ Substitute the components of $\overrightarrow a$: $\overrightarrow a - 2\widehat k = (2\widehat i + \widehat j + 2\widehat k) - 2\widehat k = 2\widehat i + \widehat j$ The square of the magnitude of this vector is: ${\left| {2\widehat i + \widehat j} \right|^2} = (2)^2 + (1)^2 + (0)^2 = 4 + 1 = 5$

Step 4: Sum the magnitudes squared of the three terms. Now, we add the results from Steps 1, 2, and 3: $S = 5 + 8 + 5$ $S = 18$

Alternative General Approach (for verification and understanding) Let $\overrightarrow a = x\widehat i + y\widehat j + z\widehat k$. Then $\widehat i \times (\overrightarrow a \times \widehat i) = \overrightarrow a - (\widehat i \cdot \overrightarrow a)\widehat i = (x\widehat i + y\widehat j + z\widehat k) - x\widehat i = y\widehat j + z\widehat k$. The magnitude squared is $|y\widehat j + z\widehat k|^2 = y^2 + z^2$.

Similarly, $\widehat j \times (\overrightarrow a \times \widehat j) = \overrightarrow a - (\widehat j \cdot \overrightarrow a)\widehat j = (x\widehat i + y\widehat j + z\widehat k) - y\widehat j = x\widehat i + z\widehat k$. The magnitude squared is $|x\widehat i + z\widehat k|^2 = x^2 + z^2$.

And, $\widehat k \times (\overrightarrow a \times \widehat k) = \overrightarrow a - (\widehat k \cdot \overrightarrow a)\widehat k = (x\widehat i + y\widehat j + z\widehat k) - z\widehat k = x\widehat i + y\widehat j$. The magnitude squared is $|x\widehat i + y\widehat j|^2 = x^2 + y^2$.

The sum is $(y^2 + z^2) + (x^2 + z^2) + (x^2 + y^2) = 2(x^2 + y^2 + z^2) = 2|\overrightarrow a|^2$.

For $\overrightarrow a = 2\widehat i + \widehat j + 2\widehat k$, we have $x=2, y=1, z=2$. $|\overrightarrow a|^2 = 2^2 + 1^2 + 2^2 = 4 + 1 + 4 = 9$. So, $2|\overrightarrow a|^2 = 2 \times 9 = 18$.

This confirms our step-by-step calculation.

Common Mistakes & Tips

• Misapplying the Vector Triple Product Identity: Ensure the order of vectors in the identity is correctly maintained. Remember it's $(\vec{A} \cdot \vec{C})\vec{B} - (\vec{A} \cdot \vec{B})\vec{C}$.
• Errors in Dot Products: Carefully calculate dot products between basis vectors and the given vector. Remember that $\vec{u} \cdot \vec{v} = |\vec{u}||\vec{v}|\cos\theta$, and for orthogonal unit vectors, this simplifies significantly.
• Magnitude Calculation: When calculating the magnitude squared of a vector expressed in components, ensure all components are squared and summed correctly.

Summary

The problem requires the evaluation of a sum of squared magnitudes of vector triple products. We utilized the vector triple product identity $\vec{A} \times (\vec{B} \times \vec{C}) = (\vec{A} \cdot \vec{C})\vec{B} - (\vec{A} \cdot \vec{B})\vec{C}$ to simplify each term. By systematically applying this identity and evaluating the necessary dot products with the given vector $\overrightarrow a = 2\widehat i + \widehat j + 2\widehat k$ and the unit vectors $\widehat i, \widehat j, \widehat k$, we found the magnitudes squared of the resulting vectors. Summing these values yielded the final answer. An alternative general approach confirmed the result by showing the expression simplifies to $2|\overrightarrow a|^2$.

The final answer is $\boxed{18}$.