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Question #73307

Difficulty: 8/10 | 🟢 Free

The Question

Let $P_1$ be a plane with normal vector $\vec{n}_1 = l\hat{i} + m\hat{j} + n\hat{k}$ , where $l, m, n$ are direction cosines such that $n = \frac{1}{2}$ . Let $P_2$ be the plane $\sqrt{3}x + y + 2\sqrt{3}z = 0$ . If $M$ is the maximum possible value of $\cos \alpha$ , where $\alpha$ is the acute angle between $P_1$ and $P_2$ , and $x \in (0, \frac{\pi}{2})$ satisfies the equation $4M^2 \sin^2 x - 4M \sin x + 1 = 0$ , then the value of $\sec^2 x$ is:

Options

A.
$\frac{3}{2}$
B.
$\frac{5}{2}$
C.
$2$
D.
$3$

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

Direction Cosines constraint|Angle between two planes|Auxiliary Angle Method

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 coupling of direction cosines and its effect on the auxiliary angle method.

Students often treat the components of a normal vector as independent variables.

When maximizing the dot product expression $\sqrt{3}l + m$ , a common trap is to blindly apply the standard auxiliary amplitude $\sqrt{a^2+b^2} = \sqrt{3+1} = 2$ , ignoring the normalization constraint $l^2 + m^2 = 1 - n^2 = 3/4$ .

This 'overflows' the valid range of the auxiliary function, leading to an incorrect maximum $M = (2+\sqrt{3})/4$ , which completely breaks the recursive trigonometric equation in Part B.

Recognizing that the domain of the auxiliary angle is restricted by the 3D geometry constraint is the key to bypassing the trap.

Quick Facts About This Question

Question Information

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