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Question #41567

Difficulty: 8/10 | 🟢 Free

The Question

Let $C$ be a closed curve in 3D space given by the parametric equations: $x(t) = 2\cos t + \sin t$ $y(t) = 2\cos t - 2\sin t$ $z(t) = \cos t + 2\sin t$ for $t \in [0, 2\pi]$ . Let $P$ be a plane passing through the origin. The direction cosines $(l, m, n)$ of the normal to the plane $P$ satisfy the equations $2l + m + 2n = 0$ and $l^2 + m^2 - n^2 = 0$ , with $l > 0$ and $m > 0$ . If $A$ is the area of the region enclosed by the orthogonal projection of the curve $C$ onto the plane $P$ , find the value of $A$ .

Options

A.
$\frac{18\sqrt{2}\pi}{5}$
B.
$6\sqrt{2}\pi$
C.
$9\pi$
D.
$6\pi$

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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

Parametric Curve Area|Scalar Multiplication Geometric Scaling|Direction Cosines constraint|Parametric form pitfalls

Why This Question Is Tricky

Common Mistakes

This question tests the 'Parametric form pitfall' and 'Direction Cosines constraint'.

Students often attempt to project the parametric equations directly onto the plane and use standard 2D area integrals, which fails because the projected basis vectors are not orthonormal.

The correct approach uses Scalar Multiplication Geometric Scaling ( $A_{proj} = A_{orig} |\cos \theta|$ ).

Furthermore, the direction cosines constraint requires careful selection of the root based on the inequality $m > 0$ , trapping students who blindly use the first algebraic solution ( $n = -l$ ) which leads to the distractor $6\sqrt{2}\pi$ .

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Application of Integrals - Class 12
Concept	Parametric Curve Area
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free