Q.Three vectors $\vec{a}$, $\vec{b}$ and $\vec{c}$ satisfy the condition $\vec{a}+\vec{b}+\vec{c}=\vec{0}$. Evaluate the quantity $\mu=\vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{c}+\vec{c}\cdot\vec{a}$, if $|\vec{a}|=3$, $|\vec{b}|=4$ and $|\vec{c}|=2$.

Key idea: square the relation $\vec{a}+\vec{b}+\vec{c}=\vec{0}$ using the dot product to link the pairwise products to the magnitudes.

Step 1 — Dot the sum with itself. Since $\vec{a}+\vec{b}+\vec{c}=\vec{0}$,

Step 2 — Expand.

Step 3 — Substitute $|\vec{a}|=3$, $|\vec{b}|=4$, $|\vec{c}|=2$. …

When three vectors add to the zero vector they form a closed triangle. We are asked for the sum of their pairwise dot products, and the cleanest route is to square the given relation — the dot product of a vector with itself is its magnitude squared, which is exactly the data we are given.

1. Start from the condition

2. Dot each side with itself

Expanding the left side and grouping the equal cross terms (since $\vec{x}\cdot\vec{y} = \vec{y}\cdot\vec{x}$):

3. Recognise $\mu$

The bracket is precisely the quantity we want:

4. Put in the magnitudes

With $|\vec{a}| = 3$, $|\vec{b}| = 4$, $|\vec{c}| = 2$: …

Method: Pairwise dot products from a zero vector sum

When vectors satisfy a linear relation such as $\vec{a}+\vec{b}+\vec{c} = \vec{0}$ and you need a combination of their pairwise dot products, square the relation with the dot product — the self-dot-products become the known magnitudes.

Steps

Step 1: Dot the given relation with itself

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

Step 2: Expand using distributivity and symmetry

$|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2 + 2(\vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{c}+\vec{c}\cdot\vec{a}) = 0.$ …

Common Mistakes

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

Why it's wrong: expanding the square gives $2(\vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{c}+\vec{c}\cdot\vec{a})$ because each pair occurs twice; writing it once makes the final value wrong by a factor of $2$. Correct approach: keep the $2$, so $|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2 + 2\mu = 0$.

Mistake 2: Trying to find each dot product individually.

Why it's wrong: the single relation only pins down the sum $\vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{c}+\vec{c}\cdot\vec{a}$, not the separate products, which are not determined. Correct approach: solve directly for the whole bracket rather than for individual terms. …