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: Pick a useful identity.
When three vectors add to the zero vector, squaring their sum gives a handy equation. The square of a vector sum expands like an algebra expression.
\[ |\vec{a} + \vec{b} + \vec{c}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}) \]

Step 2: Use the unit length fact.
Each vector is a unit vector, so each magnitude is $1$ and each squared magnitude is also $1$.

Step 3: Use the zero sum.
Since $\vec{a} + \vec{b} + \vec{c} = \vec{0}$, the left side is $0^2 = 0$.

Step 4: Substitute the values.
Put $0$ on the left and $1 + 1 + 1$ for the squared magnitudes.
\[ 0 = 3 + 2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}) \]

Step 5: Isolate the dot product sum.
Move the $3$ over.
\[ 2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}) = -3 \]

Step 6: Divide by 2.
So the wanted sum is $-\frac{3}{2}$.
\[ \boxed{-\dfrac{3}{2}} \]