Key Concepts and Formulas

• Vector Sum Identity: For any three vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$, the square of their sum's magnitude is given by: $|\vec{a} + \vec{b} + \vec{c}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a})$ This identity is derived from the distributive property of the dot product: $(\vec{u} + \vec{v}) \cdot (\vec{w} + \vec{x}) = \vec{u} \cdot \vec{w} + \vec{u} \cdot \vec{x} + \vec{v} \cdot \vec{w} + \vec{v} \cdot \vec{x}$.
• Magnitude of the Null Vector: The magnitude of the null vector $\vec{0}$ is $0$. Therefore, $|\vec{0}|^2 = 0$.
• Dot Product Property: The dot product of a vector with itself is the square of its magnitude: $\vec{v} \cdot \vec{v} = |\vec{v}|^2$.

Step-by-Step Solution

Step 1: Utilize the given vector sum. We are given that $\vec{a} + \vec{b} + \vec{c} = \vec{0}$.

• Why this step? This equation provides the fundamental relationship between the three vectors, which is essential for connecting their individual magnitudes to the sum of their pairwise dot products.

Step 2: Square the magnitude of the vector sum. We square both sides of the equation from Step 1. Since the sum of the vectors is the null vector, its magnitude is 0. $|\vec{a} + \vec{b} + \vec{c}|^2 = |\vec{0}|^2$ $|\vec{a} + \vec{b} + \vec{c}|^2 = 0$

• Why this step? Squaring the magnitude allows us to introduce terms involving the squares of individual magnitudes and the pairwise dot products, as outlined in the Vector Sum Identity.

Step 3: Expand the squared magnitude using the Vector Sum Identity. Applying the identity from the "Key Concepts and Formulas" section to the left side of the equation from Step 2: $|\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0$

• Why this step? This expansion is the core of the solution. It transforms the vector equation into an algebraic equation involving the known magnitudes and the unknown sum of dot products.

Step 4: Substitute the given magnitudes. We are given $|\vec{a}| = 1$, $|\vec{b}| = 2$, and $|\vec{c}| = 3$. Substitute these values into the expanded equation: $(1)^2 + (2)^2 + (3)^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0$

• Why this step? By substituting the specific numerical values of the magnitudes, we convert the general vector identity into a specific algebraic equation that can be solved for the desired expression.

Step 5: Solve for the expression $\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}$. First, calculate the sum of the squared magnitudes: $1 + 4 + 9 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0$ $14 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0$ Next, isolate the term containing the dot products: $2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = -14$ Finally, divide by 2 to find the value of the expression: $\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = \frac{-14}{2}$ $\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = -7$

• Why this step? This is the final algebraic manipulation to isolate and compute the value of the expression we were asked to find.

Common Mistakes & Tips

• Forgetting the factor of 2: The expansion of the squared vector sum includes a factor of 2 multiplying the sum of the pairwise dot products. Ensure this factor is not overlooked during calculations.
• Misapplying the dot product: Remember that $\vec{v} \cdot \vec{v} = |\vec{v}|^2$, not just $|\vec{v}|$.
• Algebraic errors: Double-check your arithmetic, especially when moving terms across the equals sign and when dividing.

Summary

The problem is solved by leveraging the vector sum identity, which relates the magnitude of the sum of vectors to the sum of their squared magnitudes and their pairwise dot products. Given that the sum of the vectors is the null vector, its squared magnitude is zero. By substituting the provided magnitudes into the expanded identity and performing algebraic simplification, we were able to determine the value of the sum of the pairwise dot products.

The final answer is $\boxed{-7}$.