If a, b and c are unit vectors, then |a-b|² + |b-c|² + |c-a|² does NOT exceed
(a) 4
(b) 9
(c) 8
(d)

Q: Why can't we just assume the vectors are orthogonal to maximize the distance?

That is a very tempting shortcut, but it is a trap! Assuming orthogonality is just one specific configuration. In JEE Advanced, we need the universal truth that holds for *all* possible orientations of these unit vectors. By using the algebraic method, we find the absolute upper bound that covers every possible case, not just the ones that look convenient.

Q: Why does the inequality sign flip when we multiply by -1?

Ah, a classic algebra trap! Remember that multiplying or dividing by a negative number reverses the direction of the inequality because you are essentially reflecting the values across the origin on the number line. If $x -y$. Always keep a sharp eye on that sign change—it is where many students lose marks!

Q: Is it possible for the sum to actually reach 9?

Yes, it is! It happens when $\vec{a} + \vec{b} + \vec{c} = \vec{0}$. Geometrically, this means the three vectors form a closed equilateral triangle when placed head-to-tail. This occurs when they are coplanar and separated by exactly $120^\circ$ from each other. It is a beautiful moment of geometric symmetry!

Question

If $\vec{a}, \vec{b}$ and $\vec{c}$ are unit vectors, then $|\vec{a}-\vec{b}|^2 + |\vec{b}-\vec{c}|^2 + |\vec{c}-\vec{a}|^2$ does NOT exceed

Options:

(A) $4$

(B) $9$

(C) $8$

(D) $6$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are given three unit vectors, $\vec{a}$, $\vec{b}$, and $\vec{c}$. In the context of JEE Advanced, the term "unit vectors" implies a fundamental constraint:
$$|\vec{a}| = |\vec{b}| = |\vec{c}| = 1$$

These vectors represent three explorers starting from the same origin, with their tips constrained to the surface of a unit sphere. Our objective is to find the maximum value of the expression:
$$S = |\vec{a}-\vec{b}|^2 + |\vec{b}-\vec{c}|^2 + |\vec{c}-\vec{a}|^2$$

### The Power of Expansion

Do not view $|\vec{a}-\vec{b}|^2$ merely as vector subtraction; view it as a geometric distance. In vector algebra, the squared magnitude of a difference is the key to unlocking dot products via the identity:
$$|\vec{u}-\vec{v}|^2 = |\vec{u}|^2 + |\vec{v}|^2 - 2(\vec{u} \cdot \vec{v})$$

Applying this to our first term, we obtain:
$$|\vec{a}-\vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 - 2(\vec{a} \cdot \vec{b})$$

Since $|\vec{a}| = 1$ and $|\vec{b}| = 1$, this simplifies elegantly to:
$$|\vec{a}-\vec{b}|^2 = 2 - 2(\vec{a} \cdot \vec{b})$$

### The Summation Strategy

Applying this logic to all three terms, we have:
$$|\vec{a}-\vec{b}|^2 = 2 - 2(\vec{a} \cdot \vec{b})$$
$$|\vec{b}-\vec{c}|^2 = 2 - 2(\vec{b} \cdot \vec{c})$$
$$|\vec{c}-\vec{a}|^2 = 2 - 2(\vec{c} \cdot \vec{a})$$

Summing these to find $S$, we arrive at:
$$S = 6 - 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a})$$

To maximize $S$, we must minimize the sum of the dot products: $(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a})$.

### The Master Trick: The Sum Vector

To bound the sum of dot products, we utilize the "Sum Vector" technique. Consider the vector $\vec{v} = \vec{a} + \vec{b} + \vec{c}$. Since the squared magnitude of any vector is non-negative, we have:
$$|\vec{a} + \vec{b} + \vec{c}|^2 \ge 0$$

Expanding this expression yields:
$$|\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) \ge 0$$

Substituting the unit magnitudes ($1+1+1=3$), we get:
$$3 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) \ge 0$$

### The Final Leap

Rearranging the inequality, we find:
$$2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) \ge -3$$

Multiplying by $-1$ flips the inequality:
$$-2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) \le 3$$

Returning to our expression for $S$:
$$S = 6 + [-2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a})]$$

Since the bracketed term is at most $3$, the maximum value of $S$ is:
$$S_{max} = 6 + 3 = 9$$

The **maximum value is 9**, achieved when the vectors are perfectly balanced. This technique is a classic JEE winner.

Animated Solution for Mathematics - Vector Algebra: If $a, b$ and $c$ are unit vectors, then $∣ a - b ∣^{2} + ∣ b - c ∣^{2} + ∣ c - a ∣^{2}$ does NOT exceed

Select Answer:

Visualized Solution

Visualizing the Unit Vectors

Defining the Target Expression

Expanding the First Term

Simplifying the First Term

Summing the Squared Differences

The Simplified Sum Expression

The Self-Dot Product Inequality

Expanding the Sum Vector

Bounding the Dot Product Sum

Finding the Upper Bound of S

Geometric Interpretation

The Sigma Insight: Scalar (Dot) Product

Analyzing the Setup

The Power of Expansion

The Summation Strategy

The Master Trick

The Final Leap

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Animated Solution for Mathematics - Vector Algebra: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If a,b and c are unit vectors, then ∣a−b∣2+∣b−c∣2+∣c−a∣2 does NOT exceed