Q.If $\vec{a}, \vec{b}, \vec{c}$ are unit vectors such that $\vec{a}+\vec{b}+\vec{c}=\vec{0},$ find the value of $\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}.$

Concept: Vector Magnitude Properties — using $|\vec{a}+\vec{b}+\vec{c}|^2 = 0$ for unit vectors.

Since $\vec{a}+\vec{b}+\vec{c} = \vec{0}$, its squared magnitude is zero:

Expanding:

Each vector is a unit vector, so $|\vec{a}|^2 = |\vec{b}|^2 = |\vec{c}|^2 = 1$. Substituting: …

When you see a problem involving unit vectors summing to zero, your first instinct should be: square the sum. Why? Because the dot product of a vector with itself gives its magnitude squared, and the cross terms give exactly the dot products you need. It’s a clean, algebraic way to turn a vector equation into a scalar one.

Here, $\vec{a}, \vec{b}, \vec{c}$ are unit vectors, so $|\vec{a}| = |\vec{b}| = |\vec{c}| = 1$. And they satisfy $\vec{a} + \vec{b} + \vec{c} = \vec{0}$. We want $S = \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}$.

Let’s work through it step by step.

1. Start with the given condition.

   $\vec{a} + \vec{b} + \vec{c} = \vec{0}$

2. Take the dot product of both sides with themselves.

   That is, compute $(\vec{a} + \vec{b} + \vec{c}) \cdot (\vec{a} + \vec{b} + \vec{c}) = \vec{0} \cdot \vec{0} = 0$.

3. Expand the left side using the distributive property of the dot product.

   $(\vec{a} + \vec{b} + \vec{c}) \cdot (\vec{a} + \vec{b} + \vec{c}) = \vec{a} \cdot \vec{a} + \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c} + \vec{b} \cdot \vec{a} + \vec{b} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} + \vec{c} \cdot \vec{b} + \vec{c} \cdot \vec{c}$

4. Simplify using symmetry and magnitude properties.

   Since $\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}$, etc., the nine terms reduce to:

   $|\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a})$

5. Plug in the known magnitudes.

   Each is a unit vector, so $|\vec{a}|^2 = 1$, $|\vec{b}|^2 = 1$, $|\vec{c}|^2 = 1$.

   Thus: $1 + 1 + 1 + 2S = 0$ …

Method: Squaring a Vector Sum to Extract Pairwise Dot Products

Reach for this whenever you know the magnitudes of several vectors and a linear relation among them (typically their sum) and need the sum of their pairwise dot products.

Steps

Step 1: Square the given vector equation.

From $\vec{a}+\vec{b}+\vec{c}=\vec{0}$, dot each side with itself: $|\vec{a}+\vec{b}+\vec{c}|^2 = 0$.

Step 2: Expand using the sum-of-squares identity.

$|\vec{a}+\vec{b}+\vec{c}|^2 = |\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2 + 2(\vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{c}+\vec{c}\cdot\vec{a})$ …

Common Mistakes

Mistake 1: Dropping the factor of $2$ on the cross terms.

Why it's wrong: each pair like $\vec{a}\cdot\vec{b}$ appears twice in the expansion, so the identity carries $2(\vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{c}+\vec{c}\cdot\vec{a})$; missing the $2$ gives $S=-3$. Correct approach: keep the factor, so $3+2S=0$ and $S=-\frac32$.

Mistake 2: Using $|\vec{a}|=1$ but forgetting to square. …