Q.If $\vec{a}$, $\vec{b}$, $\vec{c}$ are three vectors such that $\vec{a}+\vec{b}+\vec{c}=\vec{0}$ and $|\vec{a}|=2$, $|\vec{b}|=3$, $|\vec{c}|=5$, then value of $\vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{c}+\vec{c}\cdot\vec{a}$ is
(A) $0$
(B) $1$
(C) $-19$
(D) $38$

Concept: Use the squared magnitude of the sum of vectors — the dot‑product sum appears when expanding $|\vec{a}+\vec{b}+\vec{c}|^2$.

Since $\vec{a}+\vec{b}+\vec{c}=\vec{0}$, we have $|\vec{a}+\vec{b}+\vec{c}|^2 = 0$.

Expanding:

Substitute the given magnitudes: …

When three vectors add to zero, they form a closed triangle. The dot product sum $\vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{c}+\vec{c}\cdot\vec{a}$ is intimately linked to the squared magnitudes. Instead of guessing angles, we use a clean algebraic trick: square the vector sum.

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

2. Expand the square. The dot product distributes like ordinary multiplication (but it’s commutative for dot products). So:

Remember: $\vec{a}\cdot\vec{a} = |\vec{a}|^2$, and similarly for $\vec{b}$ and $\vec{c}$.

3. Plug in the known magnitudes. $|\vec{a}|=2$, $|\vec{b}|=3$, $|\vec{c}|=5$. So:

The equation becomes:

4. Solve for the required sum. $2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}) = -38$ …

Method: Dot-product sum from a zero-sum of vectors with known magnitudes

Use this whenever $\vec{a}+\vec{b}+\vec{c}=\vec{0}$ with $|\vec{a}|,|\vec{b}|,|\vec{c}|$ given (the vectors need not be unit vectors).

Steps

Step 1: Square the zero-sum condition

Step 2: Expand using $\vec{v}\cdot\vec{v}=|\vec{v}|^2$

Step 3: Solve for the pairwise-dot sum …

Common Mistakes

Mistake 1: Thinking the answer is $0$ because the vectors sum to zero

Why it's wrong: the zero-sum fixes directions, not the dot products; squaring gives $38+2S=0\Rightarrow S=-19$. Correct approach: expand $|\vec a+\vec b+\vec c|^2=0$ and solve.

Mistake 2: Forgetting to square the magnitudes

Why it's wrong: using $2+3+5$ instead of $2^2+3^2+5^2$ gives the wrong constant. Correct approach: $|\vec a|^2+|\vec b|^2+|\vec c|^2=4+9+25=38$. …