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Question #41931

Difficulty: 8/10 | 🟢 Free

The Question

Let $\vec{u} = -\hat{i} + \hat{j}$ and $\vec{v}(x) = (x+|x|)\hat{i} + (x-|x|)\hat{j} + 2x\hat{k}$ for $x \in \mathbb{R} \setminus \{0\}$ . Let $\alpha$ be the angle between the vectors $\vec{u}$ and $\vec{v}(x)$ when $x > 0$ , and $\beta$ be the angle between the vectors $\vec{u}$ and $\vec{v}(x)$ when $x < 0$ . Find the value of $\alpha + \beta$ .

Options

A.
$\frac{4\pi}{3}$
B.
$\pi$
C.
$\frac{2\pi}{3}$
D.
$\frac{5\pi}{3}$

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

Angle between two lines|Dot Product Projection Sign|Modulus Sign Dependency

Why This Question Is Tricky

Common Mistakes

This question tests the precise evaluation of vector magnitudes and dot products under piecewise conditions.

The primary trap lies in calculating the magnitude $|\vec{v}(x)| = \sqrt{8x^2}$ for $x < 0$ .

Students routinely simplify $\sqrt{x^2}$ as $x$ instead of $|x|$ , leading to an incorrect magnitude of $2\sqrt{2}x$ (which is negative).

This algebraic error flips the sign of $\cos \beta$ from $-1/2$ to $1/2$ , yielding an incorrect angle of $\pi/3$ and a trap answer of $\pi$ .

Furthermore, it tests the distinction between the angle between vectors (which can be obtuse) and the angle between lines.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Three-Dimensional Geometry - Class 12
Concept	Angle between two lines
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free