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Question #72356

Difficulty: 8/10 | 🟢 Free

The Question

Let $\vec{a} = (a_1, a_2, a_3)$ and $\vec{b} = (b_1, b_2, b_3)$ be non-collinear vectors in $\mathbb{R}^3$ such that $\sum_{i=1}^3 a_i^2 = p^2$ and $\sum_{i=1}^3 b_i^2 = q^2$ . For a given real parameter $r$ satisfying $r > pq$ , what is the minimum value of the expression $E = \sum_{1 \le i < j \le 3} (a_i b_j - a_j b_i)^2 + \frac{r^4}{\sum_{1 \le i < j \le 3} (a_i b_j - a_j b_i)^2}$

Options

A.
2r^2
B.
$p^2 q^2 + \frac{r^4}{p^2 q^2}$
C.
$p^2 q^2 - r^2 + \frac{r^4}{p^2 q^2}$
D.
$\frac{p^2 q^2 + r^2}{2}$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Monster

Ultra-high difficulty questions designed to test your limit, stamina, and deep conceptual understanding.

Concepts Behind This Question

Lagrange Identity|Arithmetic Mean Inequality|Closed vs Open Intervals

Why This Question Is Tricky

Examiner Levers

Variable Manipulation

Algebraic solutions only (No numbers); tests pure symbolic logic.

Difficulty Boost : +4

Common Mistakes

This question tests the critical boundary conditions of the Arithmetic Mean-Geometric Mean (AM-GM) inequality when applied to vector identities.

Students often blindly apply AM-GM to expressions of the form $x + k/x$ to get $2\sqrt{k}$ , failing to check if the equality condition ( $x = \sqrt{k}$ ) is achievable.

By using Lagrange's Identity, the domain of the variable $t = |\vec{a} \times \vec{b}|^2$ is naturally restricted to a closed upper bound $p^2 q^2$ .

Since the parameter $r$ forces the AM-GM equality point outside this interval, students must use calculus to realize the function is strictly decreasing on the restricted domain, making the boundary point the true minimum.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Vector Algebra - Class 12
Concept	Lagrange Identity
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free