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Question #40917

Difficulty: 8/10 | 🟢 Free

The Question

Let $\vec{a}, \vec{b},$ and $\vec{c}$ be unit vectors in $\mathbb{R}^3$ satisfying the equation $\vec{a} \times (\vec{b} \times \vec{c}) = \frac{1}{2} \vec{b} - \frac{1}{2} \vec{c}$ The maximum possible value of $|\vec{a} + \vec{b} + \vec{c}|^2$ is

Options

A.
7
B.
8
C.
9
D.
10

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Specialist

Targeted practice for niche, high-difficulty concepts that frequently appear in JEE Advanced exams.

Concepts Behind This Question

Collinearity and Coplanarity|Vector Triple Product Expansion|The Zero-Factor Cancellation Trap

Why This Question Is Tricky

Common Mistakes

This question tests the 'Zero-Factor Cancellation Trap' in the context of vector linear independence.

When students reach the equation $(\vec{a} \cdot \vec{c} - 1/2)\vec{b} = (\vec{a} \cdot \vec{b} - 1/2)\vec{c}$ , the algorithmic reflex is to equate the scalar coefficients to zero, implicitly assuming that $\vec{b}$ and $\vec{c}$ are linearly independent.

This leads them to a supremum of 7 (Option A).

However, they fail to check the singular case where the vectors are collinear ( $\vec{b} = \vec{c}$ ).

In this configuration, the equation collapses to $0=0$ , removing the constraints on $\vec{a}$ entirely and allowing the system to reach a higher maximum of 9.

This penalizes students who manipulate vector equations without verifying linear independence.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Introduction to Three-dimensional Geometry - Class 11
Concept	Collinearity and Coplanarity
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free