Q.If $\vec{a}$, $\vec{b}$, $\vec{c}$ are 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) $1$
(B) $3$
(C) $-\dfrac{3}{2}$
(D) None of these

The key idea is that for any vector, the dot product with itself equals the square of its magnitude. Since each vector is a unit vector, $|\vec{a}|^2 = |\vec{b}|^2 = |\vec{c}|^2 = 1$.

Step 1: Start with the given condition $\vec{a}+\vec{b}+\vec{c}=\vec{0}$. Take the dot product of each side with itself:

Step 2: Expand the left-hand side:

Step 3: Substitute $|\vec{a}|^2 = |\vec{b}|^2 = |\vec{c}|^2 = 1$: …

The key insight here is that when vectors sum to zero, they form a closed triangle. Since each vector is a unit vector, that triangle is equilateral — all sides have length 1. The dot product between any two unit vectors equals the cosine of the angle between them. In an equilateral triangle, each interior angle is $60^\circ$, but careful: the vectors are arranged head-to-tail, so the angle between any two when placed tail-to-tail is $120^\circ$, not $60^\circ$. That gives $\cos 120^\circ = -\frac12$ for each pair, and three such terms sum to $-\frac32$.

But let's prove it algebraically — no geometry needed, just the algebra of dot products.

1. Start with the given condition: $\vec{a} + \vec{b} + \vec{c} = \vec{0}$. Take the dot product of this sum with itself:

2. Expand the left side using the distributive property of the dot product:

Since the dot product is commutative ($\vec{a}\cdot\vec{b} = \vec{b}\cdot\vec{a}$), the cross terms combine:

3. Now use the fact that $\vec{a}$, $\vec{b}$, $\vec{c}$ are unit vectors. For any unit vector $\vec{u}$, $\vec{u}\cdot\vec{u} = |\vec{u}|^2 = 1$. So:

That is:

4. Solve for $S$: $2S = -3 \quad\Rightarrow\quad S = -\frac{3}{2}.$ …

Method: The "square the sum" trick for vectors adding to zero

Use this whenever vectors of known magnitude satisfy a sum condition (like $\vec{a}+\vec{b}+\vec{c}=\vec{0}$) and you need a sum of pairwise dot products.

Steps

Step 1: Take the squared magnitude of the whole sum

For any vector, $|\vec{v}|^2=\vec{v}\cdot\vec{v}$. Apply it to the given sum:

Step 2: Expand like an algebraic square

Because the dot product is distributive and commutative,

This isolates exactly the pairwise-dot sum you want.

Step 3: Substitute the constraint …

Common Mistakes

Mistake 1: Concluding the dot-product sum is $0$ because the vectors sum to $\vec 0$

Why it's wrong: $\vec a+\vec b+\vec c=\vec 0$ constrains directions, not the individual dot products; squaring shows the sum is $-\tfrac32$, not $0$. Correct approach: expand $|\vec a+\vec b+\vec c|^2=0$ and solve.

Mistake 2: Taking the angle between the vectors as $60^\circ$ …