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Question #41129

Difficulty: 8/10 | 🟢 Free

The Question

Let $S = \{(x, y, z) \in \mathbb{R}^3 : x^2 + y^2 + z^2 = 1\}$ . A function $f: S \to \mathbb{R}$ is defined such that $f(x, y, z)$ is the square of the length of the orthogonal projection of the vector $\vec{v} = x\hat{i} + y\hat{j} + z\hat{k}$ onto the plane $x+y+z=0$ . If $L$ is the total length of the curve(s) forming the pre-image $f^{-1}\left(\frac{1}{2}\right)$ , find the value of $\frac{L^2}{\pi^2}$ .

Options

A.
2
B.
4
C.
8
D.
16

Check Your Answer

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

Bijectivity and Inverse|Orthogonal projection of a line|Direction Cosines constraint

Why This Question Is Tricky

Common Mistakes

This question tests the concept of pre-images for non-injective functions defined via geometric transformations.

The domain restriction $x^2+y^2+z^2=1$ acts as the direction cosines constraint, creating a spherical manifold.

The orthogonal projection from $\mathbb{R}^3$ to a 2D plane inherently loses a dimension, making the mapping many-to-one.

The specific trap lies in solving the quadratic equation $(x+y+z)^2 = 3/2$ derived from the projection magnitude.

Students who falsely assume bijectivity or overlook the absolute value nature of the distance will miss the negative root, finding only one circle instead of two, leading them to the lethal distractor option '2'.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Relations and Functions - Class 12
Concept	Bijectivity and Inverse
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free