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Question #41192

Difficulty: 8/10 | 🟢 Free

The Question

Let $\vec{a}, \vec{b},$ and $\vec{c}$ be three unit vectors in $\mathbb{R}^3$ . The angle $\theta$ between $\vec{a}$ and $\vec{b}$ is given by $\theta = \sin^{-1}\left(\frac{12}{13}\right) + \sin^{-1}\left(\frac{4}{5}\right)$ . If $\vec{c} \cdot \vec{a} = \frac{4}{5}$ and $\vec{c} \cdot \vec{b} = \frac{36}{325}$ , what is the volume of the parallelepiped formed by $\vec{a}, \vec{b},$ and $\vec{c}$ ?

Options

A.
$0$
B.
$\frac{24\sqrt{33}}{325}$
C.
$\frac{12\sqrt{66}}{325}$
D.
$\frac{56}{65}$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Specialist

Targeted practice for niche, high-difficulty concepts that frequently appear in JEE Advanced exams.

Concepts Behind This Question

The Sum-Difference Formula Constraints|Scalar Triple Product Volume Logic|Direction Cosines constraint

Why This Question Is Tricky

Common Mistakes

This question tests the intersection of inverse trigonometric identities and 3D geometry constraints.

The primary trap lies in evaluating the sum of two inverse sines.

Students often incorrectly assume $\cos(\sin^{-1}x + \sin^{-1}y)$ is positive, ignoring that the sum of two angles in the first quadrant can be obtuse (here, $\theta > \pi/2$ ).

If the incorrect positive branch ( $\cos\theta = 33/65$ ) is used, the direction cosines constraint ( $l^2+m^2+n^2=1$ ) yields a non-zero value for $n$ , leading to the distractor volume $\frac{24\sqrt{33}}{325}$ .

The rigorous application of the sum-difference formula reveals $\cos\theta = -33/65$ , which forces $n=0$ , collapsing the parallelepiped into a plane and nullifying the volume.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Inverse Trigonometric Functions - Class 12
Concept	The Sum-Difference Formula Constraints
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free