Key Concepts and Formulas

1. Vector Triple Product Identity: For any three vectors $\overrightarrow A, \overrightarrow B, \overrightarrow C$, the vector triple product $(\overrightarrow A \times \overrightarrow B) \times \overrightarrow C$ can be expanded as: $(\overrightarrow A \times \overrightarrow B) \times \overrightarrow C = (\overrightarrow A \cdot \overrightarrow C)\overrightarrow B - (\overrightarrow B \cdot \overrightarrow C)\overrightarrow A$ This identity is crucial for simplifying expressions involving a cross product of two vectors, where one of those vectors is itself a cross product.

2. Properties of Scalar (Dot) Product:

   • Magnitude Squared: $\overrightarrow x \cdot \overrightarrow x = |\overrightarrow x|^2$. This relates the dot product of a vector with itself to its squared magnitude.
   • Orthogonality: If $\overrightarrow x \cdot \overrightarrow y = 0$, then $\overrightarrow x$ and $\overrightarrow y$ are orthogonal (perpendicular).
   • Distributivity: $\overrightarrow x \cdot (\overrightarrow y + \overrightarrow z) = \overrightarrow x \cdot \overrightarrow y + \overrightarrow x \cdot \overrightarrow z$. This allows us to expand dot products involving sums of vectors, similar to algebraic distribution.
   • Commutativity: $\overrightarrow x \cdot \overrightarrow y = \overrightarrow y \cdot \overrightarrow x$. The order of vectors in a dot product does not matter.

---

Step-by-Step Solution

1. Calculate Magnitudes and Dot Products of Base Vectors

Why this step? To simplify the complex vector expression, it's highly beneficial to first compute the magnitudes and the dot product of the given vectors $\overrightarrow a$ and $\overrightarrow b$. This often reveals special properties, such as if they are unit vectors or if they are orthogonal, which can drastically simplify subsequent calculations.

Given vectors: $\overrightarrow a = {1 \over {\sqrt {10} }}\left( {3\widehat i + \widehat k} \right) = \frac{3}{\sqrt{10}}\widehat i + 0\widehat j + \frac{1}{\sqrt{10}}\widehat k$ $\overrightarrow b = {1 \over 7}\left( {2\widehat i + 3\widehat j - 6\widehat k} \right) = \frac{2}{7}\widehat i + \frac{3}{7}\widehat j - \frac{6}{7}\widehat k$

Let's compute:

• Magnitude of $\overrightarrow a$: $|\overrightarrow a|^2 = \left(\frac{3}{\sqrt{10}}\right)^2 + \left(\frac{0}{\sqrt{10}}\right)^2 + \left(\frac{1}{\sqrt{10}}\right)^2 = \frac{9}{10} + 0 + \frac{1}{10} = \frac{10}{10} = 1$ Since $|\overrightarrow a|^2 = 1$, $\overrightarrow a$ is a unit vector. This implies $\overrightarrow a \cdot \overrightarrow a = 1$.

• Magnitude of $\overrightarrow b$: $|\overrightarrow b|^2 = \left(\frac{2}{7}\right)^2 + \left(\frac{3}{7}\right)^2 + \left(-\frac{6}{7}\right)^2 = \frac{4}{49} + \frac{9}{49} + \frac{36}{49} = \frac{49}{49} = 1$ Since $|\overrightarrow b|^2 = 1$, $\overrightarrow b$ is also a unit vector. This implies $\overrightarrow b \cdot \overrightarrow b = 1$.

• Dot product $\overrightarrow a \cdot \overrightarrow b$: $\overrightarrow a \cdot \overrightarrow b = \left(\frac{3}{\sqrt{10}}\right)\left(\frac{2}{7}\right) + \left(0\right)\left(\frac{3}{7}\right) + \left(\frac{1}{\sqrt{10}}\right)\left(-\frac{6}{7}\right)$ $\overrightarrow a \cdot \overrightarrow b = \frac{6}{7\sqrt{10}} + 0 - \frac{6}{7\sqrt{10}} = 0$ Since $\overrightarrow a \cdot \overrightarrow b = 0$, the vectors $\overrightarrow a$ and $\overrightarrow b$ are orthogonal (perpendicular) to each other.

Summary of Key Properties: $\overrightarrow a \cdot \overrightarrow a = 1$ $\overrightarrow b \cdot \overrightarrow b = 1$ $\overrightarrow a \cdot \overrightarrow b = 0$ These results are fundamental for simplifying the rest of the problem.

2. Simplify the Vector Triple Product Term

Why this step? The expression contains a vector triple product: $\left( {\overrightarrow a \times \overrightarrow b } \right) \times \left( {\overrightarrow a + 2\overrightarrow b } \right)$. Applying the vector triple product identity will transform this into a linear combination of vectors $\overrightarrow a$ and $\overrightarrow b$, making it much easier to handle.

We use the identity $(\overrightarrow A \times \overrightarrow B) \times \overrightarrow C = (\overrightarrow A \cdot \overrightarrow C)\overrightarrow B - (\overrightarrow B \cdot \overrightarrow C)\overrightarrow A$. In our case: $\overrightarrow A = \overrightarrow a$ $\overrightarrow B = \overrightarrow b$ $\overrightarrow C = \overrightarrow a + 2\overrightarrow b$

Substituting these into the identity: $\left( {\overrightarrow a \times \overrightarrow b } \right) \times \left( {\overrightarrow a + 2\overrightarrow b } \right) = \left( {\overrightarrow a \cdot (\overrightarrow a + 2\overrightarrow b)} \right)\overrightarrow b - \left( {\overrightarrow b \cdot (\overrightarrow a + 2\overrightarrow b)} \right)\overrightarrow a$

3. Evaluate the Dot Products within the Triple Product

Why this step? To complete the simplification of the vector triple product, we need to evaluate the dot products that have emerged: $\overrightarrow a \cdot (\overrightarrow a + 2\overrightarrow b)$ and $\overrightarrow b \cdot (\overrightarrow a + 2\overrightarrow b)$. This is where the results from Step 1 are critical.

Using the distributive property of the dot product and the values derived in Step 1:

• First dot product: $\overrightarrow a \cdot (\overrightarrow a + 2\overrightarrow b)$ $= \overrightarrow a \cdot \overrightarrow a + \overrightarrow a \cdot (2\overrightarrow b)$ $= \overrightarrow a \cdot \overrightarrow a + 2(\overrightarrow a \cdot \overrightarrow b)$ Substitute values: $1 + 2(0) = 1$.

• Second dot product: $\overrightarrow b \cdot (\overrightarrow a + 2\overrightarrow b)$ $= \overrightarrow b \cdot \overrightarrow a + \overrightarrow b \cdot (2\overrightarrow b)$ Using commutativity ($\overrightarrow b \cdot \overrightarrow a = \overrightarrow a \cdot \overrightarrow b$): $= \overrightarrow a \cdot \overrightarrow b + 2(\overrightarrow b \cdot \overrightarrow b)$ Substitute values: $0 + 2(1) = 2$.

Now, substitute these evaluated dot products back into the expression from Step 2: $\left( {\overrightarrow a \times \overrightarrow b } \right) \times \left( {\overrightarrow a + 2\overrightarrow b } \right) = (1)\overrightarrow b - (2)\overrightarrow a$ Thus, the simplified vector triple product is $\overrightarrow b - 2\overrightarrow a$.

4. Perform the Final Scalar (Dot) Product

Why this step? We have now simplified the complex bracketed term. The final step is to substitute this simplified term back into the original expression and perform the remaining dot product.

The original expression is $\left( {2\overrightarrow a - \overrightarrow b } \right)\left[ {\left( {\overrightarrow a \times \overrightarrow b } \right) \times \left( {\overrightarrow a + 2\overrightarrow b } \right)} \right]$. Substituting the result from Step 3: $= \left( {2\overrightarrow a - \overrightarrow b } \right) \cdot \left( {\overrightarrow b - 2\overrightarrow a } \right)$

Now, expand this dot product using the distributive property: $= (2\overrightarrow a) \cdot \overrightarrow b + (2\overrightarrow a) \cdot (-2\overrightarrow a) + (-\overrightarrow b) \cdot \overrightarrow b + (-\overrightarrow b) \cdot (-2\overrightarrow a)$ $= 2(\overrightarrow a \cdot \overrightarrow b) - 4(\overrightarrow a \cdot \overrightarrow a) - (\overrightarrow b \cdot \overrightarrow b) + 2(\overrightarrow b \cdot \overrightarrow a)$

Using the results from Step 1 ($\overrightarrow a \cdot \overrightarrow a = 1$, $\overrightarrow b \cdot \overrightarrow b = 1$, $\overrightarrow a \cdot \overrightarrow b = 0$) and commutativity ($\overrightarrow b \cdot \overrightarrow a = \overrightarrow a \cdot \overrightarrow b$): $= 2(0) - 4(1) - (1) + 2(0)$ $= 0 - 4 - 1 + 0$ $= -5$

---

Common Mistakes & Tips

• Order of Operations in Vector Triple Product: Ensure you correctly identify $\overrightarrow A, \overrightarrow B, \overrightarrow C$ in the identity $(\overrightarrow A \times \overrightarrow B) \times \overrightarrow C = (\overrightarrow A \cdot \overrightarrow C)\overrightarrow B - (\overrightarrow B \cdot \overrightarrow C)\overrightarrow A$. A common error is mixing up the roles of the vectors or the terms in the expansion.
• Forgetting Orthogonality: The fact that $\overrightarrow a$ and $\overrightarrow b$ are orthogonal is a significant simplification. Always check for this condition early on by calculating the dot product.
• Algebraic Errors: Be meticulous with signs and coefficients when expanding dot products, especially when dealing with negative signs or scalar multiples.

Summary

The problem involves simplifying a complex vector expression. The strategy employed was to first calculate the magnitudes and dot product of the base vectors, which revealed them to be unit vectors and orthogonal. This greatly simplified the subsequent steps. The vector triple product was then expanded using the standard identity, and the resulting dot products were evaluated using the pre-calculated values. Finally, the remaining dot product was computed, yielding the solution.

The value of the expression $\left( {2\overrightarrow a - \overrightarrow b } \right)\left[ {\left( {\overrightarrow a \times \overrightarrow b } \right) \times \left( {\overrightarrow a + 2\overrightarrow b } \right)} \right]$ is $-5$.

The final answer is $\boxed{-5}$.