Problem 72 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Let \(\vec{a}, \vec{b}\) and \(\vec{c}\) be three unit vectors such that \(\vec{a} \times(\vec{b} \times \vec{c})=\frac{\sqrt{3}}{2}(\vec{b}+\vec{c})\). If \(\vec{b}\) is not parallel to \(\vec{c}\), then \(\Rightarrow\) the angle between a and \(b\) is: (a) \(\frac{2 \pi}{3}\) (b) \(\frac{5 \pi}{6}\) (c) \(\frac{3 \pi}{4}\) (d) \(\frac{\pi}{2}\)

Step-by-Step Solution

Answer

The angle between \(\vec{a}\) and \(\vec{b}\) is \(\frac{5\pi}{6}\).

Step 1: Using Vector Triple Product Identity

We start by using the vector triple product identity: \[\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}.\] Let's substitute this into the given equation: \[(\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c} = \frac{\sqrt{3}}{2}(\vec{b} + \vec{c}).\]

Step 2: Equating Components

We separate the equation into components of \(\vec{b}\) and \(\vec{c}\):\[(\vec{a} \cdot \vec{c})\vec{b} = \frac{\sqrt{3}}{2}\vec{b},\] \[- (\vec{a} \cdot \vec{b})\vec{c} = \frac{\sqrt{3}}{2}\vec{c}.\] This gives us two scalar equations:\(\vec{a} \cdot \vec{c} = \frac{\sqrt{3}}{2}\) and \(\vec{a} \cdot \vec{b} = -\frac{\sqrt{3}}{2}\).

Step 3: Finding the Angle Between Vectors

The dot product is related to the angle between vectors: \(\vec{a} \cdot \vec{b} = \|\vec{a}\| \|\vec{b}\| \cos \theta\), where \(\theta\) is the angle between them. Since \(\vec{a}\) and \(\vec{b}\) are unit vectors, \(\|\vec{a}\| = \|\vec{b}\| = 1\), thus:\[\cos \theta = \vec{a} \cdot \vec{b} = -\frac{\sqrt{3}}{2}.\]We find that \(\theta = \frac{5\pi}{6}\).

Key Concepts

Unit Vectors

Unit vectors are foundational in vector mathematics as they designate direction with a magnitude of exactly one. These vectors are typically denoted with a hat symbol, such as \( \hat{i} \), \( \hat{j} \), and \( \hat{k} \) for the standard basis in 3D space.
Unit vectors simplify the process of scaling and direction determination in vector equations. They are crucial when you need to identify the pure direction of a vector without influence from its magnitude. For example, a vector \( \vec{v} \) could be expressed as \( \vec{v} = \|\vec{v}\| \hat{u} \), where \( \hat{u} \) is the unit vector indicating direction and \( \|\vec{v}\| \) is the magnitude of \( \vec{v} \).

• Every vector in vector space can be reduced to a unit vector by dividing it by its magnitude.
• Unit vectors are often used to normalize vectors in both physics and mathematics to simplify problems that focus on direction.
• In the exercise, \( \vec{a} \), \( \vec{b} \), and \( \vec{c} \) are unit vectors which means that the operations and calculations focus solely on angular relationships and directions rather than magnitudes.

Vector Cross Product

The vector cross product, also known as the vector product, is a binary operation on two vectors in three-dimensional space. The result is a vector that is perpendicular to both of the original vectors involved in the operation. The cross product is defined as: \[ \vec{a} \times \vec{b} = \|\vec{a}\| \|\vec{b}\| \sin(\theta) \hat{n} \] where \( \theta \) is the angle between \( \vec{a} \) and \( \vec{b} \) and \( \hat{n} \) is a unit vector perpendicular to both \( \vec{a} \) and \( \vec{b} \).
The cross product is significant because it helps determine a new vector that is orthogonal to the original plane established by two vectors. It is vital in physics and engineering, especially in calculations involving rotational forces and torque.

• Only applicable in three-dimensional space, and result in vectors that are orthogonal to the vectors being multiplied.
• Useful for calculating moments about a point and understanding rotational dynamics.
• In this exercise, using the cross product allowed the application of the vector triple product identity to find relationships between the vectors \( \vec{b}, \vec{c}, \) and the resultant expression.

Dot Product Angle Calculation

The dot product, also known as the scalar product, is a method of multiplying two vectors that results in a scalar quantity. It is defined mathematically as: \[ \vec{a} \cdot \vec{b} = \|\vec{a}\| \|\vec{b}\| \cos(\theta) \] where \( \theta \) is the angle between the two vectors. The dot product is pivotal in determining the angle between two vectors, as well as in projecting one vector onto another.
To find the angle between two vectors, rearrange the formula to solve for \( \cos(\theta) \): \[ \cos(\theta) = \frac{\vec{a} \cdot \vec{b}}{\|\vec{a}\| \|\vec{b}\|} \] For unit vectors, this simplifies further as both magnitudes are 1, thus \( \cos(\theta) = \vec{a} \cdot \vec{b} \).

• Crucial for computing angles in vector tables and projections in physics.
• The calculated angle provides insights into vector alignment and direction.
• In the problem, the calculation of the dot product allowed you to find the angle between \( \vec{a} \) and \( \vec{b} \), which turned out to be \( \frac{5\pi}{6} \).