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Question #76005

Difficulty: 8/10 | 🟢 Free

The Question

Let a⃗\vec{a} and b⃗\vec{b} be two unit vectors such that the angle between them is θ\theta, where θ∈(π2,π)\theta \in \left(\frac{\pi}{2}, \pi\right). Let P1P_1 be the plane passing through the origin, containing the vector a⃗\vec{a}, and perpendicular to the plane containing a⃗\vec{a} and b⃗\vec{b}. Let P2P_2 be the plane passing through the origin, containing the vector b⃗\vec{b}, and perpendicular to the plane containing a⃗\vec{a} and b⃗\vec{b}. If α\alpha is the acute angle between the planes P1P_1 and P2P_2, then cos⁡α\cos \alpha is equal to:

Options

  • A.

    −cos⁡θ-\cos \theta

  • B.

    cos⁡θ\cos \theta

  • C.

    −sin⁡θ-\sin \theta

  • D.

    sin⁡θ\sin \theta

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

Vector Triple Product Expansion|Angle between two planes|Symmetric vs Non-symmetric form|ASTC Quadrant Sign Convention

Why This Question Is Tricky

Examiner Levers

Variable Manipulation

Algebraic solutions only (No numbers); tests pure symbolic logic.

Difficulty Boost : +4

Common Mistakes

This question tests the intersection of Vector Triple Product (VTP) expansions and 3D geometry, specifically targeting the definition of the angle between two planes.

Students must first correctly apply the VTP to find the normal vectors of planes defined in non-symmetric form.

The primary trap lies in the ASTC quadrant sign convention combined with the definition of the acute angle between planes.

The dot product of the normals yields cos⁡ϕ=cos⁡θ\cos \phi = \cos \theta.

Because θ∈(π/2,π)\theta \in (\pi/2, \pi), cos⁡θ\cos \theta is negative, meaning the angle between the normals is obtuse.

Students who blindly equate the angle between normals to the angle between planes will incorrectly choose cos⁡θ\cos \theta.

The true acute angle α\alpha requires taking the absolute value, yielding cos⁡α=−cos⁡θ\cos \alpha = -\cos \theta.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterVector Algebra - Class 12
ConceptVector Triple Product Expansion
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree