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Question #41859

Difficulty: 8/10 | 🟢 Free

The Question

Let $\vec{u}, \vec{v}, \vec{w} \in \mathbb{R}^3$ be three vectors such that $|\vec{u}| \le 2$ , $|\vec{v}| \le 3$ , and $|\vec{w}| \le 4$ . It is given that $\vec{u} \cdot \vec{v} \ge 3\sqrt{3}$ and $\vec{v} \cdot \vec{w} \ge 6\sqrt{2}$ . If the matrix $M = [\vec{u}\ \vec{v}\ \vec{w}]$ is invertible, let $S$ be the set of all possible values of the dot product $\vec{u} \cdot \vec{w}$ . Which of the following represents the exact set $S$ ?

Options

A.
$(2\sqrt{6} - 2\sqrt{2}, 8)$
B.
$[2\sqrt{6} - 2\sqrt{2}, 8]$
C.
$(2\sqrt{6} - 2\sqrt{2}, 8]$
D.
$[2\sqrt{6} - 2\sqrt{2}, 8)$

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Bridge

Practice problems that bridge the gap between foundational theory and advanced practical application.

Concepts Behind This Question

System of Inequalities|Dot Product Projection Sign|Linear Independence vs Dependence

Why This Question Is Tricky

Common Mistakes

This question tests the intersection of algebraic inequalities and geometric linear independence.

Students typically use the dot product formulas and angle addition properties to find the minimum and maximum possible values of $\vec{u} \cdot \vec{w}$ , arriving at the boundary values $2\sqrt{6} - 2\sqrt{2}$ and $8$ .

The critical trap lies in the condition that matrix $M$ is invertible, which dictates that the vectors must form a linearly independent basis in $\mathbb{R}^3$ .

Achieving the boundary values requires the vectors to be either collinear (for the maximum) or coplanar (for the minimum).

Failing to recognize that linear independence strictly forbids these geometric states leads students to incorrectly choose the closed interval.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Linear Inequalities - Class 11
Concept	System of Inequalities
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free