Problem 9 - Master Resource Book in JEE Main Physics

Question

Two vectors a and \(\mathbf{b}\) are at an angle of \(60^{\circ}\) with each other. Their resultant makes an angle of \(45^{\circ}\) with a. If \(|\mathbf{b}|=2\) units, then \(|\mathbf{a}|\) is (a) \(\sqrt{3}\) (b) \(\sqrt{3}-1\) (c) \(\sqrt{3}+1\) (d) \(\sqrt{3} / 2\)

Step-by-Step Solution

Answer

|\mathbf{a}| = \sqrt{3} - 1

Step 1: Understand the Problem

We have two vectors, \( \mathbf{a} \) and \( \mathbf{b} \), at an angle of \( 60^{\circ} \) with each other. The resultant vector \( \mathbf{R} \) makes an angle of \( 45^{\circ} \) with \( \mathbf{a} \), and \( |\mathbf{b}| = 2 \) units. We need to find \( |\mathbf{a}| \).

Step 2: Use Vector Addition Formula

The magnitude of the resultant vector \( \mathbf{R} \) of two vectors \( \mathbf{a} \) and \( \mathbf{b} \) making an angle \( \theta \) is given by: \[ |\mathbf{R}|^2 = |\mathbf{a}|^2 + |\mathbf{b}|^2 + 2 |\mathbf{a}| |\mathbf{b}| \cos\theta \]Substituting \( \theta = 60^{\circ} \), we have: \[ |\mathbf{R}|^2 = |\mathbf{a}|^2 + 4 + 2 \cdot 2 \cdot |\mathbf{a}| \cdot \frac{1}{2} \]Simplifying, \[ |\mathbf{R}|^2 = |\mathbf{a}|^2 + 4 + 2|\mathbf{a}| \]

Step 3: Use Relationship with Resultant Angle

We know that the resultant \( \mathbf{R} \) makes an angle \( 45^{\circ} \) with \( \mathbf{a} \). Using the formula for the tangent of the angle between the resultant and the vector \( \mathbf{a} \):\[ \tan 45^{\circ} = \frac{|\mathbf{b}| \sin\theta}{|\mathbf{a}| + |\mathbf{b}| \cos\theta} \]Substituting the given values, \[ 1 = \frac{2 \cdot \sin 60^{\circ}}{|\mathbf{a}| + 2 \cdot \cos 60^{\circ}} \]\[ 1 = \frac{2 \cdot \frac{\sqrt{3}}{2}}{|\mathbf{a}| + 1} \]\[ 1 = \frac{\sqrt{3}}{|\mathbf{a}| + 1} \]Solving this equation for \( |\mathbf{a}| \), we have: \[ |\mathbf{a}| + 1 = \sqrt{3} \]

Step 4: Solve for |a|

From the equation \( |\mathbf{a}| + 1 = \sqrt{3} \), solve for \( |\mathbf{a}| \):\[ |\mathbf{a}| = \sqrt{3} - 1 \]

Step 5: Verify the Answer

Verify if \( |\mathbf{a}| \) satisfies all conditions set by the problem. We determined \( |\mathbf{a}| = \sqrt{3} - 1 \) by solving the angular relationship, and this matches one of the given options. Therefore, this is the correct solution.

Key Concepts

Resultant Vector

In the context of vectors, a resultant vector is the single vector that has the same effect as the combined action of two or more vectors acting together. To find the resultant of two vectors, we need to consider both their magnitude and direction. Here, vectors \( \mathbf{a} \) and \( \mathbf{b} \) are at an angle of \( 60^{\circ} \) with each other. The resultant vector \( \mathbf{R} \) can be found using vector addition. This involves summing the effects of \( \mathbf{a} \) and \( \mathbf{b} \) in a way that considers how they interact based on their directions.

When two vectors are added, their magnitudes and directions determine the resultant’s magnitude and exact direction. This process often involves trigonometry to account for non-right-angle interactions. By applying the vector addition formula, the combined effect is calculated, leading to a single resultant vector that simplifies the analysis of forces or movements.

Angle Between Vectors

The angle between two vectors is crucial in determining the resultant vector’s direction and magnitude. In this problem, vector \( \mathbf{a} \) and vector \( \mathbf{b} \) form an angle of \( 60^{\circ} \). This angle affects how directly or indirectly the vectors contribute to the resultant vector.

The angle helps us use trigonometric functions like cosine and sine in calculations. The cosine of the angle specifies how much two vectors point in the same general direction, which affects their combined strength in a perpendicular direction. Understanding this angle and its role allows us to accurately compute the effects of each vector in contributing to the resultant vector positioning.

Magnitude of Vectors

The magnitude of a vector determines its strength or size and is represented by its length in a vector diagram. In this particular problem, we have \(|\mathbf{b}| = 2\) units, which is a given. We are tasked with determining \(|\mathbf{a}|\).

When calculating the magnitude, the resultant vector's formula is essential:

• Using \(|\mathbf{R}|^2 = |\mathbf{a}|^2 + |\mathbf{b}|^2 + 2 |\mathbf{a}| |\mathbf{b}| \cos\theta \),
• where \(\theta = 60^{\circ}\), the relationship becomes a mathematical expression that lets us solve for \(|\mathbf{a}|\).

In this problem, solving involves substituting known values, then algebraically manipulating the resultant equation. Thus, understanding vector magnitudes is fundamental to problem-solving involving vector addition.

Trigonometric Relationships in Vectors

Trigonometry plays a crucial role in vector calculations, especially when dealing with angles and magnitudes. Here, the tangent relationship utilizes the angles to establish a connection between the vectors and their resultant. The formula used is

• \( \tan \alpha = \frac{|\mathbf{b}| \sin\theta}{|\mathbf{a}| + |\mathbf{b}| \cos\theta} \)

where \( \alpha \) is the angle between the resultant and vector \( \mathbf{a} \). For this exercise, \( \alpha = 45^{\circ} \), leading to a

• \( \tan 45^{\circ} = 1 \)

This translates into manipulating the sets of trigonometric identities to calculate \(|\mathbf{a}| \). The use of trigonometry here includes understanding how sine and cosine functions describe vector component images, affecting how forces or movements between vectors are calculated. This builds the foundation for solving complex vector-related problems by enabling the navigation between angles and lengths efficiently.