Problem 94 - A Complete Resource Book in Mathematics for JEE Main

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 \([\mathbf{2 0 0 7}]\) (A) exactly two values of \(\theta\) (B) more than two values of \(\theta\) (C) no value of \(\bar{\theta}\) (D) exactly one value of \(\theta\)

Step-by-Step Solution

Answer

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

Step 1: Understand the Given Problem

We are given two unit vectors, \(\hat{u}\) and \(\hat{v}\), and the goal is to determine when the vector \(2 \hat{u} \times 3 \hat{v}\) is a unit vector. Our task is to find the possible values of the acute angle \(\theta\) between \(\hat{u}\) and \(\hat{v}\) that make this product a unit vector.

Step 2: Use Cross Product Formula

The magnitude of the cross product of two vectors \(\mathbf{a}\) and \(\mathbf{b}\) is given by \(\| \mathbf{a} \times \mathbf{b} \| = \| \mathbf{a} \| \cdot \| \mathbf{b} \| \cdot \sin(\theta)\). Since \(\hat{u}\) and \(\hat{v}\) are unit vectors, we have \(\|2 \hat{u}\| = 2\) and \(\|3 \hat{v}\| = 3\). The magnitude \(\|2 \hat{u} \times 3 \hat{v}\|\) becomes \(6 \sin(\theta)\).

Step 3: Set the Cross Product Magnitude Equal to 1

For \(2 \hat{u} \times 3 \hat{v}\) to be a unit vector, \(\|2 \hat{u} \times 3 \hat{v}\|\) must be equal to 1. Therefore, we set up the equation: \(6 \sin(\theta) = 1\).

Step 4: Solve for \(\theta\)

Solve the equation \(6 \sin(\theta) = 1\), which simplifies to \(\sin(\theta) = \frac{1}{6}\). The angle \(\theta\) can be calculated using \(\theta = \sin^{-1}\left(\frac{1}{6}\right)\).

Step 5: Determine Number of Solutions

Since \(\theta\) is an acute angle, it must satisfy \(0 < \theta < 90^{\circ}\). The equation \(\sin(\theta) = \frac{1}{6}\) can have exactly one acute angle solution in this range.

Key Concepts

Unit Vectors

In vector mathematics, a unit vector is a vector with a magnitude of 1. Unit vectors are often used to indicate direction. They are particularly useful in both physics and engineering for representing directions without concern for length. If a vector \(\mathbf{a}\) has a magnitude larger than 1, it can be converted to a unit vector by dividing each component by the vector's magnitude. Practically, this means finding a direction without any scale. For any vector \(\mathbf{a}\), the unit vector \(\mathbf{\hat{a}}\) is calculated as follows:

• First, find the magnitude of \(\mathbf{a}\), noted as \(\|\mathbf{a}\|\).
• Then, divide each component of the vector by its magnitude, leading to the unit vector \(\mathbf{\hat{a}} = \frac{\mathbf{a}}{\|\mathbf{a}\|}\).

In our problem, the vectors \(\hat{u}\) and \(\hat{v}\) are already unit vectors, hence they already have a magnitude of 1, simplifying calculations needed for operations such as cross products.

Acute Angle

An acute angle is any angle less than 90 degrees. This concept is crucial when working with vectors, especially in determining the nature of their relative direction. In our exercise, the angle \(\theta\) between unit vectors \(\hat{u}\) and \(\hat{v}\) must be acute, indicating their directions form a narrow intersection in space.
Using trigonometric principles, an important property to understand is that the sine of \(\theta\), noted as \(\sin(\theta)\), is always positive for acute angles. This property is vital for calculations such as determining when the cross product of vectors results in a unit vector. Solving for \(\sin(\theta) = \frac{1}{6}\) verifies \(\theta\) remains within the acute range, as \(\sin^{-1}(\frac{1}{6})\) results in a positive, acute angle.

Magnitude

The magnitude of a vector provides the "length" or "size" of a vector. It’s computed from the components of the vector using the Pythagorean theorem in its dimensional space.
For instance, given a vector \(\mathbf{a} = (a_1, a_2, a_3)\) in three-dimensional space, the magnitude is calculated as \(\|\mathbf{a}\| = \sqrt{a_1^2 + a_2^2 + a_3^2}\).
In the context of the exercise, we are interested in ensuring the magnitude of the resulting cross product of vector calculations, \(\|2 \hat{u} \times 3 \hat{v}\|\), is equal to 1. The magnitude of a cross product is determined by the formula: \(\| \mathbf{a} \times \mathbf{b} \| = \| \mathbf{a} \| \cdot \| \mathbf{b} \| \cdot \sin(\theta)\).
When calculating \(6 \sin(\theta)\) to be equal to 1, and hence a magnitude of 1, ensures the vector produced remains a unit vector. Understanding these principles allows students to manipulate vectors and understand their sizes effectively.