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) $38$

> [!TLDR] > The key idea is that squaring the zero-sum condition $\vec{a}+\vec{b}+\vec{c}=\vec{0}$ gives a direct relation between the sum of dot products and the sum of squares of magnitudes. The value is $\boxed{-19}$. 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. $$(\vec{a}+\vec{b}+\vec{c})\cdot(\vec{a}+\vec{b}+\vec{c}) = \vec{0}\cdot\vec{0} = 0$$ 2. **Expand the square.** The dot product distributes like ordinary multiplication (but it’s commutative for dot products). So: $$(\vec{a}+\vec{b}+\vec{c})\cdot(\vec{a}+\vec{b}+\vec{c}) = \vec{a}\cdot\vec{a} + \vec{b}\cdot\vec{b} + \vec{c}\cdot\vec{c} + 2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a})$$ 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: $$|\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 = 4 + 9 + 25 = 38$$ The equation becomes: $$38 + 2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}) = 0$$ 4. **Solve for the required sum.** $$2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}) = -38$$ …

If veca, vecb, vecc are three vectors such that veca+vecb+vecc=vec0 and |veca|=2, |vecb|=3, |vecc|=5, then val

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Concept understanding — Unit Vector Dot Product Sum

Sum of Dot Products of Unit Vectors

A classic Class-12 result asks: if several unit vectors add up to the zero vector, what is the sum of their pairwise dot products? The trick is always the same — square the sum — and it turns a hard-looking problem into one line of algebra.

The Master Move: Square the Magnitude

For any vector $\vec{v}$ , $|\vec{v}|^2 = \vec{v}\cdot\vec{v}$ . Applying this to a sum expands like an ordinary algebraic square, because the dot product distributes and is commutative:

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

That isolates exactly the quantity we want — the sum of the pairwise dot products.

The Standard Result

Suppose $\vec{a}, \vec{b}, \vec{c}$ are unit vectors (so each squared length is $1$ ) and $\vec{a}+\vec{b}+\vec{c}=\vec{0}$ . Then the left side is $|\vec{0}|^2=0$ , giving

0 = 1+1+1 + 2\big(\vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{c}+\vec{c}\cdot\vec{a}\big).

Solving:

\vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{c}+\vec{c}\cdot\vec{a} = -\frac{3}{2}

Why the Method Always Works

The whole technique rests on two facts: $|\vec{v}|^2=\vec{v}\cdot\vec{v}$ turns a magnitude condition into dot products, and for a unit vector $\vec{v}\cdot\vec{v}=1$ . Whatever constraint you are given (the vectors sum to zero, or to a known vector), squaring both sides produces an equation in the unknown dot-product sum.

A Variation

If instead $|\vec{a}+\vec{b}+\vec{c}| = k$ with the same three unit vectors, the same expansion gives

$\vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{c}+\vec{c}\cdot\vec{a} = \frac{k^2-3}{2}.$ …

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The key idea is that squaring the zero-sum condition $\vec{a}+\vec{b}+\vec{c}=\vec{0}$ gives a direct relation between the sum of dot products and the sum of squares of magnitudes. The value is $\boxed{-19}$ .

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.

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

(\vec{a}+\vec{b}+\vec{c})\cdot(\vec{a}+\vec{b}+\vec{c}) = \vec{0}\cdot\vec{0} = 0

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

(\vec{a}+\vec{b}+\vec{c})\cdot(\vec{a}+\vec{b}+\vec{c}) = \vec{a}\cdot\vec{a} + \vec{b}\cdot\vec{b} + \vec{c}\cdot\vec{c} + 2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a})

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

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

|\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 = 4 + 9 + 25 = 38

The equation becomes:

38 + 2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}) = 0

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

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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 understanding — Unit Vector Dot Product Sum

Sum of Dot Products of Unit Vectors

The Master Move: Square the Magnitude

The Standard Result

Why the Method Always Works

A Variation

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

Steps

Common Mistakes

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