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Question #40854

Difficulty: 8/10 | 🟢 Free

The Question

Let $L_t$ be a variable line in 3D space. The orthogonal projection of the origin $O(0,0,0)$ onto $L_t$ is the point $Q(t) = (x(t), y(t), z(t))$ , where for $t > 0$ : $x(t) = \int_0^t (2u \sin u + u^2 \cos u) du$ $y(t) = \int_0^t (2u \cos u - u^2 \sin u) du$ $z(t) = \int_0^t 3u^2 du$ Find the minimum distance between the point $P\left(0, 0, \frac{5}{3}\right)$ and the projection point $Q(t)$ .

Options

A.
$\frac{\sqrt{13}}{3}$
B.
$\frac{\sqrt{14}}{3}$
C.
$\frac{5}{3}$
D.
$\frac{\sqrt{10}}{3}$

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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

Distance Formula in 3D|Orthogonal projection of a line|Integrating f(x) + xf'(x) forms

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to recognize the reverse product rule in integration, specifically the $\int (f(x) + xf'(x)) dx = xf(x)$ identity, disguised within trigonometric functions.

A common trap is attempting integration by parts manually, which consumes excessive time and often leads to algebraic errors.

Furthermore, the problem uses geometric flavor text ('orthogonal projection onto a variable line') as a distractor; students might incorrectly attempt to find the equation of the line $L_t$ instead of realizing the coordinates of the projection $Q(t)$ are already explicitly given.

Finally, it requires setting up a distance function with a variable limit and correctly differentiating a polynomial to find the global minimum.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Introduction to Three-dimensional Geometry - Class 11
Concept	Distance Formula in 3D
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free