a, b, c are 3 vectors, such that a + b + c = 0, |a| = 1, |b| = 2, |c| = 3, then a.b + b.c + c.a is e

Q: Why can't we just use the triangle law of vector addition?

While the triangle law is a beautiful geometric concept, it is a specific case. The algebraic method of squaring the sum is a universal tool that works even when vectors are not collinear. In this problem, the vectors are collinear because $|\vec{a}| + |\vec{b}| = |\vec{c}|$, but the algebraic approach is more robust and less prone to visualization errors.

Q: Why do we square the vector sum?

We square the sum because the dot product of a vector with itself is the square of its magnitude, i.e., $\vec{x} \cdot \vec{x} = |\vec{x}|^2$. Squaring the sum allows us to bridge the gap between vector addition and the scalar dot products we are looking for. It transforms a vector equation into a scalar equation.

Q: Is the negative sign in the final answer correct?

Absolutely! A negative dot product simply means the angle between the vectors is obtuse (greater than $90^\circ$). In a system where vectors sum to zero, the dot products often reflect this 'pulling' against each other, leading to negative values.

Question

$\vec{a}, \vec{b}, \vec{c}$ are 3 vectors, such that $\vec{a} + \vec{b} + \vec{c} = 0$, $|\vec{a}| = 1, |\vec{b}| = 2, |\vec{c}| = 3$, then $\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}$ is equal to

Options:

(A) 1

(B) 0

(C) -7

(D) 7

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler of the JEE landscape. Today, we are not just solving a problem; we are exploring the hidden geometry of vectors.

We are given three vectors, $\vec{a}$, $\vec{b}$, and $\vec{c}$, with magnitudes $1$, $2$, and $3$ respectively, and a condition that their sum is zero: $\vec{a} + \vec{b} + \vec{c} = 0$. This is a state of perfect equilibrium.

Imagine these vectors as forces acting on a point; if they sum to zero, the point remains unmoved. But what does this tell us about their interactions?

### The Geometry of the Zero Sum

Before we rush into the algebra, let's pause and visualize. If you place these vectors head-to-tail, they form a closed loop.

But look at the magnitudes: $1 + 2 = 3$. This is a special case! In a typical triangle, the sum of any two sides must be greater than the third.

Here, the sum of two sides equals the third. This means our 'triangle' has collapsed into a straight line; they are collinear. While this is a fascinating geometric insight, we must be careful. Relying solely on geometry can sometimes lead to traps, so let's trust the algebra to guide us safely to the answer.

### The Algebraic Bridge

We need to find the sum of the pairwise dot products: $\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}$. How do we extract these from a simple addition?

The secret lies in the power of the dot product. We take our equation $\vec{a} + \vec{b} + \vec{c} = 0$ and dot it with itself.

This is equivalent to squaring the vector sum: 
$$(\vec{a} + \vec{b} + \vec{c}) \cdot (\vec{a} + \vec{b} + \vec{c}) = 0 \cdot 0 = 0$$
This is our bridge from the vector world to the scalar world.

### The Expansion: Unveiling the Structure

Now, let's expand this expression using the distributive property of the dot product. It behaves just like the algebraic identity $(x + y + z)^2 = x^2 + y^2 + z^2 + 2(xy + yz + zx)$.

When we expand our dot product, we get:

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

Don't let the length of this equation intimidate you. It is a beautiful, symmetric structure. We have the sum of the squares of the magnitudes, and we have the sum of the pairwise dot products, all tied together by the number two.

### The Final Calculation

Now, we simply substitute the known magnitudes: $|\vec{a}| = 1$, $|\vec{b}| = 2$, and $|\vec{c}| = 3$. Plugging these into our equation, we get:

$$(1)^2 + (2)^2 + (3)^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0$$

Calculating the squares, we have $1 + 4 + 9 = 14$. So, the equation becomes:

$$14 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0$$

Now, we isolate our target expression. Move $14$ to the other side, and it becomes $-14$:

$$2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = -14$$

Finally, divide by $2$ to find the value:

$$\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = -7$$

### The Takeaway

We have arrived at our destination: **$-7$**. This problem is a classic, and the technique of squaring the vector sum is a weapon you should always keep in your arsenal.

Remember, for any set of vectors that sum to zero, the sum of their pairwise dot products is always $-\frac{1}{2}$ of the sum of their squared magnitudes. Keep this in mind, and you will breeze through similar problems in your exams. You have mastered the logic; now go forth and conquer!

Animated Solution for Mathematics - Vector Algebra: $a, b, c$ are 3 vectors, such that $a + b + c = 0$ , $∣ a ∣ = 1, ∣ b ∣ = 2, ∣ c ∣ = 3$ , then $a \cdot b + b \cdot c + c \cdot a$ is equal to

Select Answer:

Visualized Solution

Given Vectors and Magnitudes

The Zero Sum Condition

Squaring the Vector Sum

Expanding the Dot Product

Substituting Known Values

Evaluating the Squares

Summing the Constants

Transposing the Constant

Final Division

Conclusion

The Sigma Insight: Scalar (Dot) Product

Analyzing the Setup

The Geometry of the Zero Sum

The Algebraic Bridge

The Expansion

The Final Calculation

The Takeaway

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