1. Key Concepts and Formulas

• Dot Product Property: For any vectors $\vec{u}$ and $\vec{v}$, $\vec{u} \cdot \vec{v} = \vec{v} \cdot \vec{u}$.
• Magnitude Squared: For any vector $\vec{v}$, $\vec{v} \cdot \vec{v} = |\vec{v}|^2$.
• Vector Sum Squared: The square of the sum of three vectors $(\vec{a} + \vec{b} + \vec{c})$ is given by the dot product of the sum with itself: $(\vec{a} + \vec{b} + \vec{c}) \cdot (\vec{a} + \vec{b} + \vec{c}) = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a})$ This is analogous to the algebraic identity $(x+y+z)^2 = x^2+y^2+z^2+2(xy+yz+zx)$.

2. Step-by-Step Solution

Step 1: Start with the given vector sum. We are given that the sum of the three vectors is the zero vector: $\overrightarrow a + \overrightarrow b + \overrightarrow c = \overrightarrow 0$

• Reasoning: This is the primary condition provided in the problem, and it's the starting point for all subsequent manipulations.

Step 2: Square both sides of the vector equation. To introduce terms involving magnitudes and dot products, we take the dot product of both sides of the equation with themselves. This is often referred to as "squaring" the vector equation: $(\overrightarrow a + \overrightarrow b + \overrightarrow c) \cdot (\overrightarrow a + \overrightarrow b + \overrightarrow c) = \overrightarrow 0 \cdot \overrightarrow 0$ We know that $\vec{v} \cdot \vec{v} = |\vec{v}|^2$, and for the zero vector, $|\overrightarrow 0|^2 = 0$. Therefore, the equation becomes: ${\left( {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right)^2} = 0$

• Reasoning: Squaring the vector sum allows us to utilize the distributive property of the dot product and relate the sum of vectors to the individual magnitudes and pairwise dot products, which are central to the problem's objective.

Step 3: Expand the squared vector sum. Using the formula for the square of a vector sum (as stated in Key Concepts), we expand the left side of the equation: ${\left| {\overrightarrow a } \right|^2} + {\left| {\overrightarrow b } \right|^2} + {\left| {\overrightarrow c } \right|^2} + 2\left( {\overrightarrow a \,.\,\overrightarrow b + \overrightarrow b \,.\,\overrightarrow c + \overrightarrow c \,.\,\overrightarrow a } \right) = 0$

• Reasoning: This expansion explicitly generates the term $\left( {\overrightarrow a \,.\,\overrightarrow b + \overrightarrow b \,.\,\overrightarrow c + \overrightarrow c \,.\,\overrightarrow a } \right)$, which is what we need to find. The terms involving the squared magnitudes are also present, and their values are given.

Step 4: Substitute the given magnitudes. The problem provides the magnitudes of the individual vectors: ${\left| {\overrightarrow a } \right| = 5},\quad {\left| {\overrightarrow b } \right| = 4},\quad {\left| {\overrightarrow c } \right| = 3}$ Substitute these values into the expanded equation from Step 3: ${(5)^2} + {(4)^2} + {(3)^2} + 2\left( {\overrightarrow a \,.\,\overrightarrow b + \overrightarrow b \,.\,\overrightarrow c + \overrightarrow c \,.\,\overrightarrow a } \right) = 0$

• Reasoning: This step incorporates the specific numerical information given in the problem, allowing us to transform the vector equation into an algebraic equation with known numerical values.

Step 5: Simplify and solve for the sum of dot products. Calculate the squares of the magnitudes and sum them: $25 + 16 + 9 + 2\left( {\overrightarrow a \,.\,\overrightarrow b + \overrightarrow b \,.\,\overrightarrow c + \overrightarrow c \,.\,\overrightarrow a } \right) = 0$ $50 + 2\left( {\overrightarrow a \,.\,\overrightarrow b + \overrightarrow b \,.\,\overrightarrow c + \overrightarrow c \,.\,\overrightarrow a } \right) = 0$ Now, isolate the term containing the sum of dot products: $2\left( {\overrightarrow a \,.\,\overrightarrow b + \overrightarrow b \,.\,\overrightarrow c + \overrightarrow c \,.\,\overrightarrow a } \right) = -50$ Divide by 2: $\left( {\overrightarrow a \,.\,\overrightarrow b + \overrightarrow b \,.\,\overrightarrow c + \overrightarrow c \,.\,\overrightarrow a } \right) = \frac{-50}{2}$ $\left( {\overrightarrow a \,.\,\overrightarrow b + \overrightarrow b \,.\,\overrightarrow c + \overrightarrow c \,.\,\overrightarrow a } \right) = -25$

• Reasoning: This is the algebraic manipulation to find the precise value of the expression $\overrightarrow a \,.\,\overrightarrow b + \overrightarrow b \,.\,\overrightarrow c + \overrightarrow c \,.\,\overrightarrow a$.

Step 6: Calculate the absolute value. The question asks for the value of $\left| {\overrightarrow a .\overrightarrow b + \overrightarrow b .\overrightarrow c + \overrightarrow c .\overrightarrow a } \right|$. We have found the value of the expression inside the absolute value in Step 5. Now, we take its absolute value: $\left| {\overrightarrow a \,.\,\overrightarrow b + \overrightarrow b \,.\,\overrightarrow c + \overrightarrow c \,.\,\overrightarrow a } \right| = \left| -25 \right|$ $\left| {\overrightarrow a \,.\,\overrightarrow b + \overrightarrow b \,.\,\overrightarrow c + \overrightarrow c \,.\,\overrightarrow a } \right| = 25$

• Reasoning: This final step directly addresses the requirement of the question, which is to find the absolute value of the sum of the dot products.

3. Common Mistakes & Tips

• Forgetting the Absolute Value: The question specifically asks for the absolute value. A common error is to stop at the value $-25$ and not take the absolute value.
• Incorrect Expansion: Ensure the expansion of $(\vec{a} + \vec{b} + \vec{c})^2$ is done correctly, including the $2 \times$ factor for the cross-terms.
• Geometric Interpretation: The condition $\vec{a} + \vec{b} + \vec{c} = \vec{0}$ means that the vectors form a closed triangle when placed head-to-tail. The magnitudes 3, 4, 5 form a Pythagorean triple, indicating these vectors could form a right-angled triangle, but this geometric property is not directly needed for this algebraic solution method.

4. Summary

The problem is solved by leveraging the vector identity for squaring a sum of vectors. Given $\vec{a} + \vec{b} + \vec{c} = \vec{0}$, squaring this equation leads to $|\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0$. Substituting the given magnitudes and solving for the expression in parentheses yields $-25$. Taking the absolute value of this result gives the final answer.

The final answer is $\boxed{25}$.