If a, b and c are unit vectors satisfying |a - b|² + |b - c|² + |c - a|² = 9, then |2a + 5b + 5c| is

Q: Wait, why does $|\vec{a} - \vec{b}|^2$ expand to $|\vec{a}|^2 + |\vec{b}|^2 - 2\vec{a} \cdot \vec{b}$? Isn't it just $|\vec{a}|^2 + |\vec{b}|^2$?

That is a classic trap! In scalar algebra, $(a-b)^2 = a^2 + b^2 - 2ab$. However, in vector algebra, the square of a difference involves the dot product because vectors have both magnitude and direction. The term $-2\vec{a} \cdot \vec{b}$ is essential because it accounts for the angle between the vectors. Without it, you are ignoring the geometric relationship between them.

Q: If $|\vec{a} + \vec{b} + \vec{c}|^2 = 0$, does it automatically mean $\vec{a} + \vec{b} + \vec{c} = \vec{0}$?

Yes, absolutely. The magnitude of a vector is a non-negative scalar. If the square of the magnitude is zero, the magnitude itself must be zero. The only vector that has a magnitude of zero is the null vector, $\vec{0}$. This is a powerful conclusion that simplifies the entire problem.

Q: Can I solve this by assuming $\vec{a}, \vec{b}, \vec{c}$ are specific vectors like $(1,0,0)$?

While you might get lucky with specific values, it is risky and not recommended for JEE Advanced. The problem is general, and the algebraic approach using dot products is the robust, intended method. It ensures your logic holds true for any configuration of these vectors, not just the ones you happen to pick.

Question

If $\vec{a}, \vec{b}$ and $\vec{c}$ are unit vectors satisfying $|\vec{a} - \vec{b}|^2 + |\vec{b} - \vec{c}|^2 + |\vec{c} - \vec{a}|^2 = 9$, then $|2\vec{a} + 5\vec{b} + 5\vec{c}|$ is .........

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a field, and you have three unit vectors, $\vec{a}$, $\vec{b}$, and $\vec{c}$, all originating from the same point. The problem states they are unit vectors, which means their lengths are exactly one: $|\vec{a}| = |\vec{b}| = |\vec{c}| = 1$.

Geometrically, you can visualize these vectors as three arrows pointing from the center of a unit sphere to its surface. We are given the constraint:
$$|\vec{a} - \vec{b}|^2 + |\vec{b} - \vec{c}|^2 + |\vec{c} - \vec{a}|^2 = 9$$

This equation is a statement about the geometric balance of these vectors.

### The Algebraic Expansion

To unlock this, we expand the squared differences using the fundamental identity $|\vec{x} - \vec{y}|^2 = |\vec{x}|^2 + |\vec{y}|^2 - 2\vec{x} \cdot \vec{y}$. Applying this to our given equation, we obtain:

$$(|\vec{a}|^2 + |\vec{b}|^2 - 2\vec{a} \cdot \vec{b}) + (|\vec{b}|^2 + |\vec{c}|^2 - 2\vec{b} \cdot \vec{c}) + (|\vec{c}|^2 + |\vec{a}|^2 - 2\vec{c} \cdot \vec{a}) = 9$$

Since each vector magnitude is one, the equation simplifies significantly:

$$2(1 + 1 + 1) - 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 9$$

This leads to the expression:

$$6 - 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 9$$

Solving for the sum of the dot products, we find:

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

### The Revelation

We now utilize the identity for the square of a sum:
$$|\vec{a} + \vec{b} + \vec{c}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a})$$

Substituting our known values into this identity:

$$|\vec{a} + \vec{b} + \vec{c}|^2 = 3 + 2\left(-\frac{3}{2}\right) = 3 - 3 = 0$$

This is the critical "Aha!" moment. Because the magnitude of the sum is zero, we conclude that $\vec{a} + \vec{b} + \vec{c} = \vec{0}$.

### The Final Victory

We are tasked with finding the value of $|2\vec{a} + 5\vec{b} + 5\vec{c}|$. Since $\vec{a} + \vec{b} + \vec{c} = \vec{0}$, we can substitute $\vec{b} + \vec{c} = -\vec{a}$ into the expression:

$$|2\vec{a} + 5(\vec{b} + \vec{c})| = |2\vec{a} + 5(-\vec{a})|$$

Simplifying the vector arithmetic:

$$|2\vec{a} - 5\vec{a}| = |-3\vec{a}| = 3|\vec{a}|$$

Given that $|\vec{a}| = 1$, the final magnitude is:

**3**

Animated Solution for Mathematics - Vector Algebra: If $a, b$ and $c$ are unit vectors satisfying $∣ a - b ∣^{2} + ∣ b - c ∣^{2} + ∣ c - a ∣^{2} = 9$ , then $∣2 a + 5 b + 5 c ∣$ is .........

Enter Numerical Value:

Visualized Solution

Defining Unit Vectors

Expanding the Given Equation

Substituting Unit Magnitudes

Solving for Dot Product Sum

Relating to $∣ a + b + c ∣^{2}$

Calculating the Sum Magnitude

Deducing $a + b + c = 0$

Simplifying the Target Expression

Final Calculation

Key Takeaways

The Sigma Insight: Scalar (Dot) Product

Analyzing the Setup

The Algebraic Expansion

The Revelation

The Final Victory

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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 satisfying ∣a−b∣2+∣b−c∣2+∣c−a∣2=9, then ∣2a+5b+5c∣ is .........