JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

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 statements:

(1) $|\vec{a} + \lambda\vec{c}| \ge |\vec{a}|$ for all $\lambda \in \mathbb{R}$ .

(2) $\vec{a}$ and $\vec{c}$ are always parallel.

Then:

Choose the correct answer:

Explanation

Solution

1. Simplify the given condition:

|\vec{a} + \vec{b} + \vec{c}|^2 = |\vec{a} + \vec{b} - \vec{c}|^2

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

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

Since $\vec{b} \cdot \vec{c} = 0$ , we conclude that $\vec{a} \cdot \vec{c} = 0$ . This means $\vec{a}$ and $\vec{c}$ are perpendicular, not parallel. Statement (2) is incorrect.

2. Evaluate Statement (1):

|\vec{a} + \lambda\vec{c}|^2 = |\vec{a}|^2 + \lambda^2|\vec{c}|^2 + 2\lambda(\vec{a} \cdot \vec{c})

Since $\vec{a} \cdot \vec{c} = 0$ :

|\vec{a} + \lambda\vec{c}|^2 = |\vec{a}|^2 + \lambda^2|\vec{c}|^2

Since $\lambda^2|\vec{c}|^2 \ge 0$ , it follows that $|\vec{a} + \lambda\vec{c}|^2 \ge |\vec{a}|^2$ , or $|\vec{a} + \lambda\vec{c}| \ge |\vec{a}|$ . Statement (1) is correct.

Correct Option: (2)

Choose the correct answer:

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