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Question #41774

Difficulty: 8/10 | 🟢 Free

The Question

Let $\vec{a}$ and $\vec{b}$ be two mutually orthogonal unit vectors in $\mathbb{R}^3$ . Let $\vec{r}$ be a variable vector such that $|\vec{r}| = \sqrt{3}$ . Consider the objective function $f(\vec{r}) = \frac{1}{3} \vec{r} \cdot \left( (\vec{r} \times \vec{a}) \times (\vec{r} \times \vec{b}) \right) + (\vec{r} \times \vec{a}) \cdot (\vec{r} \times \vec{b})$ The maximum value of $f(\vec{r})$ is

Options

A.
2
B.
$\sqrt{3}$
C.
$\frac{3}{2}$
D.
$\frac{5}{2}$

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

Objective Function Gradient|Vector Triple Product Expansion|Lagrange Identity

Why This Question Is Tricky

Common Mistakes

The problem frames the objective function as a complex vector expression to test the simultaneous application of the Vector Triple Product and Lagrange's Identity.

The core trap lies in the transition from the vector field to the scalar quadratic form $z - xy$ .

Students frequently decouple the variables, erroneously assuming that $z$ and the product $-xy$ can be maximized independently (yielding $\sqrt{3} + 1.5$ ), or they incorrectly apply AM-GM without respecting the spherical domain constraint $x^2+y^2+z^2=3$ .

By forcing the substitution $u = x^2+y^2$ , the problem demands rigorous single-variable calculus to find the true geometric maximum, penalizing those who ignore the interdependent bounds of the components.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Linear Programming - Class 12
Concept	Objective Function Gradient
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free