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Question #41735

Difficulty: 8/10 | 🟢 Free

The Question

Let a curve $C$ be defined parametrically by $x = t+1$ , $y = 2|t-1|$ , and $z = t^2 - 3t + 3$ for $t \in \mathbb{R}$ . Find the minimum distance from a point on the curve $C$ to the plane $x + y + z = 2$ .

Options

A.
$\frac{\sqrt{3}}{3}$
B.
$\frac{4\sqrt{3}}{3}$
C.
$\frac{2\sqrt{3}}{3}$
D.
0

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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

Distance from point to plane|Critical Point vs Stationary Point|Non-Differentiable Extrema

Why This Question Is Tricky

Common Mistakes

This question tests the distinction between stationary points ( $f'(x)=0$ ) and critical points where $f'(x)$ does not exist.

By embedding an absolute value within a distance minimization problem, students are lured into blindly differentiating the piecewise functions.

Doing so yields stationary points ( $t=0$ and $t=2$ ) that lie outside their respective domain branches, leading to a false minimum distance of $4/\sqrt{3}$ .

The true global minimum lies at the non-differentiable cusp ( $t=1$ ), forcing students to rely on algebraic bounding rather than standard differential calculus.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Three-Dimensional Geometry - Class 12
Concept	Distance from point to plane
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free