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Question #75001

Difficulty: 8/10 | 🟢 Free

The Question

Let the direction cosines $(l, m, n)$ of two lines $L_1$ and $L_2$ satisfy the equations $2l - m + 2n = 0$ and $mn + nl + lm = 0$ . Let $\theta$ be the acute angle between $L_1$ and $L_2$ , and define $\alpha = \sin^2\theta$ . If $(l_1, m_1, n_1)$ and $(l_2, m_2, n_2)$ are the direction cosines of $L_1$ and $L_2$ respectively, and $S = |m_1 n_2 - n_1 m_2| + |n_1 l_2 - l_1 n_2| + |l_1 m_2 - m_1 l_2|$ then the value of $9S^2 + \alpha$ is:

Options

A.
26
B.
10
C.
28
D.
25

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

Direction Cosines constraint|Lagrange Identity Equivalents|Modulus Function Symmetry

Why This Question Is Tricky

Examiner Levers

Recursive Logic

The answer to part A is required to solve part B (Step-wise failure test).

Difficulty Boost : +5

Common Mistakes

This question tests the application of Lagrange's Identity in 3D geometry.

Students often memorize that $\sum (m_1 n_2 - n_1 m_2)^2 = \sin^2\theta$ .

When faced with the sum of the moduli of these components, a common trap is to either ignore the modulus and compute the algebraic sum squared (yielding 10), or incorrectly apply the Cauchy-Schwarz equality condition $S^2 = 3\sin^2\theta$ (yielding 28), assuming the components are symmetric.

By forcing the calculation of unequal components, the modulus trap catches students who bypass exact component derivation.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Three-Dimensional Geometry - Class 12
Concept	Direction Cosines constraint
Difficulty	8
Cognitive Load	7
Target Solve Time	170 sec
Access	Free