Problem 89 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If \(\hat{u}\) and \(\hat{v}\) are unit vectors and \(\theta\) is the acute angle between them, then \(2 \hat{u} \times 3 \hat{v}\) is a unit vector for \(\quad[2007]\) (a) no value of \(\theta\) (b) exactly one value of \(\theta\) (c) exactly two values of \(\theta\) (d) more than two values of \(\theta\)

Step-by-Step Solution

Answer

(b) exactly one value of \(\theta\)

Step 1: Determine the Cross Product Magnitude Formula

First, recall that the magnitude of the cross product of two vectors is given by the formula:\[|\hat{u} \times \hat{v}| = |\hat{u}||\hat{v}|\sin\theta\]Since \(\hat{u}\) and \(\hat{v}\) are unit vectors, \(|\hat{u}| = 1\) and \(|\hat{v}| = 1\), simplifying the magnitude to:\[|\hat{u} \times \hat{v}| = \sin\theta\]

Step 2: Scale the Cross Product by Given Constants

Calculate the magnitude of the scaled cross product \(2\hat{u} \times 3\hat{v}\). This becomes:\[|2\hat{u} \times 3\hat{v}| = 2 \times 3 \times |\hat{u} \times \hat{v}| = 6 \sin\theta\]

Step 3: Set the Condition for Unit Vector

For \(2 \hat{u} \times 3 \hat{v}\) to be a unit vector, its magnitude must be 1. Thus, we set up the following equation:\[6 \sin \theta = 1\]Solving for \(\sin \theta\) gives:\[\sin \theta = \frac{1}{6}\]

Step 4: Determine Valid Values of \(\theta\)

Since \(\theta\) is an acute angle, it must satisfy \(0 < \theta < \frac{\pi}{2}\). The equation \(\sin \theta = \frac{1}{6}\) has exactly one solution in the interval \((0, \frac{\pi}{2})\).

Key Concepts

Unit Vectors

Unit vectors are vectors with a magnitude of one. They are important in vector calculus because they provide direction information without altering scale. A unit vector is typically represented with a hat symbol, such as \( \hat{u} \) or \( \hat{v} \).
To find a unit vector in the same direction as any given vector \( \mathbf{a} \), you divide the vector by its magnitude. If \( \mathbf{a} = \langle x, y, z \rangle \), then the unit vector \( \hat{a} \) is given by:

• \( \hat{a} = \frac{\mathbf{a}}{||\mathbf{a}||} = \left\langle \frac{x}{||\mathbf{a}||}, \frac{y}{||\mathbf{a}||}, \frac{z}{||\mathbf{a}||} \right\rangle \)

Unit vectors are often used to represent directions in physics and engineering. They ensure that the direction does not depend on the magnitude, staying within a normalized range of one.

Cross Product

The cross product is an operation on two vectors in three-dimensional space, producing a third vector that is perpendicular to both of the original vectors. The result of a cross product is often crucial in determining perpendicularity and areas. If you have two vectors \( \mathbf{a} = \langle a_1, a_2, a_3 \rangle \) and \( \mathbf{b} = \langle b_1, b_2, b_3 \rangle \), their cross product \( \mathbf{a} \times \mathbf{b} \) is given by:

• \( \mathbf{a} \times \mathbf{b} = \langle a_2b_3 - a_3b_2, a_3b_1 - a_1b_3, a_1b_2 - a_2b_1 \rangle \)

In the exercise, the cross product of unit vectors \( \hat{u} \times \hat{v} \) returns a vector with a magnitude of \( \sin \theta \), where \( \theta \) is the angle between \( \hat{u} \) and \( \hat{v} \). This concept shows how cross products are not just about direction but also about magnitude.

Acute Angle

An acute angle is any angle that is less than 90 degrees or equivalently, less than \( \frac{\pi}{2} \) radians. When dealing with vectors, angles between them can tell us how they are oriented relative to each other.
In the context of the original exercise, finding the acute angle \( \theta \) where \( \sin \theta = \frac{1}{6} \) is important because it determines when the magnitude of the cross product equals one, making the cross product a unit vector.
To find this angle, you would typically consult trigonometric tables or use an inverse sine function.

Magnitude Formula

The magnitude formula is vital in determining the size or length of a vector. In vector calculus, it's often used alongside operations like cross and dot products. The magnitude of a vector \( \mathbf{v} = \langle x, y, z \rangle \) is calculated as:

• \( ||\mathbf{v}|| = \sqrt{x^2 + y^2 + z^2} \)

For unit vectors, the magnitude is always one. But when you have vectors scaled by constants, like in \( 2\hat{u} \times 3\hat{v} \), the magnitude changes accordingly. The magnitude in the given exercise is \( 6 \sin \theta \), because the constants multiply the unit cross product result of \( \sin \theta \).
Thus, setting \( 6 \sin \theta = 1 \) allowed us to solve for the angle that ensures a unit vector result, showing an application of how magnitude plays into vector scaling.