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Question #40879

Difficulty: 8/10 | 🟢 Free

The Question

Let $S$ be the set of points $P(x,y,z)$ in $\mathbb{R}^3$ satisfying the equations $x+y+z=1$ and $|x|+|y|+|z|=1$ . Let $A$ be the point $(1, 1, 1)$ and $B$ be the point $(-1, 2, 2)$ . The minimum value of $PA + PB$ as $P$ varies over the region $S$ is:

Options

A.
$\frac{\sqrt{102}}{3}$
B.
$\frac{\sqrt{6} + \sqrt{22}}{2}$
C.
$\frac{2\sqrt{3} + \sqrt{66}}{3}$
D.
$\sqrt{2} + \sqrt{6}$

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

Octant Sign Convention|Angle between line and plane|Triangle Inequality Mind Games

Why This Question Is Tricky

Common Mistakes

This problem tests the 'Octant Sign Convention' and 'Triangle Inequality' traps.

The condition $|x|+|y|+|z|=1$ combined with $x+y+z=1$ restricts the domain to the first octant.

Students who mechanically apply the reflection principle to minimize $PA+PB$ will find the unconstrained minimum on the plane, leading to a point outside the allowed octant (violating the modulus constraint).

Recognizing this forces a boundary optimization, testing the student's ability to integrate algebraic constraints with 3D geometric feasibility.

Quick Facts About This Question

Question Information

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