Let a = 2i + j + k, and b and c be two nonzero vectors such that |a + b + c| = |a + b - c| and b · c

Q: Why do we square the magnitude? Can't we just take the square root?

Taking the square root of a vector magnitude equation is mathematically undefined because vectors do not have a 'sign' in the scalar sense. Squaring is the standard, rigorous technique to convert the geometric modulus into a scalar dot product, which is the language of algebra.

Q: Why group $(\vec{a} + \vec{b})$? Isn't it easier to expand all three terms?

Expanding three terms like $|\vec{a} + \vec{b} + \vec{c}|^2$ leads to a massive expression with six dot product terms. By grouping, we treat $(\vec{a} + \vec{b})$ as a single vector $\vec{u}$, reducing the complexity to a simple binomial expansion. It is about working smarter, not harder.

Q: Why is Statement (A) true? How does $\vec{a} \perp \vec{c}$ prove it?

Since we proved $\vec{a} \perp \vec{c}$, the vectors $\vec{a}$ and $\lambda \vec{c}$ form the legs of a right-angled triangle. The resultant vector $\vec{a} + \lambda \vec{c}$ is the hypotenuse. By the Pythagorean theorem, the square of the hypotenuse is $|\vec{a}|^2 + |\lambda \vec{c}|^2$. Since $|\lambda \vec{c}|^2 \geq 0$, the hypotenuse is always greater than or equal to the base $|\vec{a}|$, making Statement (A) correct.

Question

Let $\vec{a} = 2\hat{i} + \hat{j} + \hat{k}$, and $\vec{b}$ and $\vec{c}$ be two nonzero vectors such that $|\vec{a} + \vec{b} + \vec{c}| = |\vec{a} + \vec{b} - \vec{c}|$ and $\vec{b} \cdot \vec{c} = 0$. Consider the following two statement:
(A) $|\vec{a} + \lambda \vec{c}| \geq |\vec{a}|$ for all $\lambda \in \mathbb{R}$.
(B) $\vec{a}$ and $\vec{c}$ are always parallel.

Options:

(A) only (B) is correct

(B) neither (A) nor (B) is correct

(C) only (A) is correct

(D) both (A) and (B) are correct

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow explorers of the JEE Advanced frontier. Today, we are not just solving a problem; we are uncovering the hidden geometric architecture of vectors.

We are given a vector $\vec{a} = 2\hat{i} + \hat{j} + \hat{k}$, and two mysterious non-zero vectors $\vec{b}$ and $\vec{c}$. We are armed with two critical clues: $|\vec{a} + \vec{b} + \vec{c}| = |\vec{a} + \vec{b} - \vec{c}|$ and the orthogonality condition $\vec{b} \cdot \vec{c} = 0$.

### The Power of Squaring

When you face a magnitude equation, do not panic. The modulus is a geometric concept, but the dot product is an algebraic one.

To bridge this gap, we square both sides:
$$|\vec{a} + \vec{b} + \vec{c}|^2 = |\vec{a} + \vec{b} - \vec{c}|^2$$

This is our first masterstroke. By squaring, we transform the modulus into the dot product of the vector with itself, using the identity $|\vec{u}|^2 = \vec{u} \cdot \vec{u}$.

### The Art of Grouping

Expanding three terms directly is a recipe for algebraic chaos. Instead, we use a strategic grouping.

Let us treat $(\vec{a} + \vec{b})$ as a single vector unit, say $\vec{u}$. Now, our equation looks like $|\vec{u} + \vec{c}|^2 = |\vec{u} - \vec{c}|^2$.

Expanding this using the binomial identity $|\vec{x} \pm \vec{y}|^2 = |\vec{x}|^2 + |\vec{y}|^2 \pm 2(\vec{x} \cdot \vec{y})$, we get:
$$|\vec{u}|^2 + |\vec{c}|^2 + 2(\vec{u} \cdot \vec{c}) = |\vec{u}|^2 + |\vec{c}|^2 - 2(\vec{u} \cdot \vec{c})$$

Look at the elegance of this moment! The terms $|\vec{u}|^2$ and $|\vec{c}|^2$ appear on both sides. They cancel out, leaving us with $2(\vec{u} \cdot \vec{c}) = -2(\vec{u} \cdot \vec{c})$.

Bringing everything to one side, we find $4(\vec{u} \cdot \vec{c}) = 0$, which simplifies to $(\vec{a} + \vec{b}) \cdot \vec{c} = 0$.

### The Orthogonality Discovery

Now, we distribute the dot product: $\vec{a} \cdot \vec{c} + \vec{b} \cdot \vec{c} = 0$. We know from our initial conditions that $\vec{b} \cdot \vec{c} = 0$.

Substituting this, we arrive at the beautiful, simple truth: $\vec{a} \cdot \vec{c} = 0$. This means $\vec{a}$ and $\vec{c}$ are perfectly perpendicular.

### Evaluating the Statements

With $\vec{a} \perp \vec{c}$, Statement (A) becomes a question of geometry. The vector $\vec{a} + \lambda \vec{c}$ is the hypotenuse of a right-angled triangle with legs $\vec{a}$ and $\lambda \vec{c}$.

By Pythagoras, we have:
$$|\vec{a} + \lambda \vec{c}|^2 = |\vec{a}|^2 + |\lambda \vec{c}|^2$$

Since $|\lambda \vec{c}|^2$ is always non-negative, $|\vec{a} + \lambda \vec{c}|^2 \geq |\vec{a}|^2$, which confirms Statement (A) is correct.

Statement (B) claims $\vec{a}$ and $\vec{c}$ are parallel, but we have proven they are perpendicular. Thus, Statement (B) is incorrect. We have conquered the problem, not by brute force, but by understanding the soul of the vectors.

Select Answer:

Visualized Solution

Given Conditions

Squaring the Magnitude

Grouping Terms

Expanding the Left Side

Expanding the Right Side

Canceling Common Terms

Simplifying the Equation

Distributing the Dot Product

Using the Second Condition

Conclusion of Orthogonality

Evaluating Statement (A)

Visualizing the Resultant

Applying Pythagoras Theorem

Evaluating Statement (B)

Final Conclusion

The Sigma Insight: Scalar (Dot) Product

Analyzing the Setup

The Power of Squaring

The Art of Grouping

The Orthogonality Discovery

Evaluating the Statements

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