Question

If $\vec{a}$, $\vec{b}$, and $\vec{c}$ are three unit vectors such that $\vec{a} + \vec{b} + \vec{c} = \vec{0}$, then the value of $\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}$ is:

• A. $-\frac{3}{2}$
• B. $\frac{3}{2}$
• C. $0$
• D. $-3$

Hint:

If the sum of $n$ unit vectors is $\vec{0}$, the sum of their pairwise dot products is always $-\frac{n}{2}$. Here, $n=3$, so $-\frac{3}{2}$.

Answer 1

The Correct Option is A

Step 1: Concept
We use the algebraic vector expansion formula for the square of the sum of three vectors: \[ |\vec{a} + \vec{b} + \vec{c}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) \]

Step 2: Meaning
Since $\vec{a}$, $\vec{b}$, and $\vec{c}$ are unit vectors, their magnitudes are $1$. The sum of the vectors is given as $\vec{0}$.

Step 3: Analysis
Substitute the magnitudes and the sum vector into the expansion formula: \[ 0^2 = 1^2 + 1^2 + 1^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) \] \[ 0 = 3 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) \] \[ 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = -3 \] \[ \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = -\frac{3}{2} \]

Step 4: Conclusion
The value of the expression is $-\frac{3}{2}$.

Final Answer: (A)