Problem 44 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Let \(\vec{a}\) and \(\vec{b}\) be two unit vectors. If the vectors \(\vec{c}=\hat{a}+2 \hat{b}\) and \(\vec{d}=5 \hat{a}-4 \hat{b}\) are perpendicular to each other, then the angle between \(\hat{a}\) and \(\hat{b}\) is : (a) \(\frac{\pi}{6}\) (b) \(\frac{\pi}{2}\) (c) \(\frac{\pi}{3}\) (d) \(\frac{\pi}{4}\)

Step-by-Step Solution

Answer

The angle between \(\vec{a}\) and \(\vec{b}\) is \(\frac{\pi}{3}\). (Answer: c)

Step 1: Understand the Problem

We are given two unit vectors \(\vec{a}\) and \(\vec{b}\). Vector \(\vec{c}\) is defined as \(\vec{a} + 2\vec{b}\) and vector \(\vec{d}\) as \(5\vec{a} - 4\vec{b}\). These vectors are perpendicular, and we need to find the angle between \(\vec{a}\) and \(\vec{b}\).

Step 2: Use the Perpendicular Condition

If vectors \(\vec{c}\) and \(\vec{d}\) are perpendicular, then their dot product must be zero. Thus, \((\vec{a} + 2\vec{b}) \cdot (5\vec{a} - 4\vec{b}) = 0\).

Step 3: Expand the Dot Product

Expand the expression: \[ (\vec{a} + 2\vec{b}) \cdot (5\vec{a} - 4\vec{b}) = \vec{a} \cdot 5\vec{a} + \vec{a} \cdot (-4\vec{b}) + 2\vec{b} \cdot 5\vec{a} + 2\vec{b} \cdot (-4\vec{b}) \]This simplifies to: \[ 5(\vec{a}\cdot\vec{a}) - 4(\vec{a}\cdot\vec{b}) + 10(\vec{b}\cdot\vec{a}) - 8(\vec{b}\cdot\vec{b}) = 0 \]

Step 4: Simplify Using Unit Vector Properties

Since \(\vec{a}\) and \(\vec{b}\) are unit vectors, \(\vec{a}\cdot\vec{a} = 1\) and \(\vec{b}\cdot\vec{b} = 1\). Replace these in the equation:\[ 5(1) - 4(\vec{a}\cdot\vec{b}) + 10(\vec{b}\cdot\vec{a}) - 8(1) = 0 \]Simplify to get:\[ -3 + 6(\vec{a}\cdot\vec{b}) = 0 \]

Step 5: Solve for the Dot Product

Solve the equation for \(\vec{a}\cdot\vec{b}\):\[ 6(\vec{a}\cdot\vec{b}) = 3 \]\[ \vec{a}\cdot\vec{b} = \frac{1}{2} \]

Step 6: Find the Angle

Recall the dot product formula \(\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos\theta\). Since both vectors are unit vectors, \(|\vec{a}| = 1\) and \(|\vec{b}| = 1\). Therefore, \(\cos\theta = \frac{1}{2}\).The angle \(\theta\) whose cosine is \(\frac{1}{2}\) is \(\frac{\pi}{3}\).

Key Concepts

Understanding Unit Vectors

Unit vectors are the building blocks of vector calculus. They have a magnitude of exactly one. This simplifies many calculations because when you work with unit vectors, their magnitudes don't affect the math. In notation, unit vectors are often denoted with a hat, like \( \hat{a} \) or \( \hat{b} \).
These vectors maintain their direction but serve more as directional guides without scaling the vector's inherent value. Understanding unit vectors is crucial because they help in simplifying vector operations and making more complex vector manipulations easier. For instance:

• When you find the dot product, the simplification \( \vec{a} \cdot \vec{a} = 1 \) applies.
• Working with unit vectors often leads to discovering angles between components more readily.

Therefore, whenever vectors are defined as unit vectors, any operation in vector calculus, like finding angles or projections, becomes straightforward since you don't have to factor in magnitude.

Grasping the Dot Product

The dot product is an operation that takes two vectors and returns a scalar, summarizing the extent to which the vectors point in the same direction. Mathematically, for vectors \( \vec{a} \) and \( \vec{b} \), it’s expressed as \( \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos\theta \).
In essence, it is the product of the magnitudes of the two vectors and the cosine of the angle between them. Through this operation, we deduce important information about the spatial relationship between the vectors.

• If the dot product is positive, the vectors point in a similar direction.
• If the dot product is zero, they are perpendicular (orthogonal).
• If negative, they point in opposite directions.

This property is vital. For example, understanding that perpendicular vectors (like \( \vec{c} \) and \( \vec{d} \) in our problem) have a dot product of zero helps solve geometry and physics problems efficiently.

Calculating the Angle Between Vectors

Finding the angle between vectors is a fundamental operation in vector calculus, illustrating their directional relationship. When dealing with unit vectors \( \vec{a} \) and \( \vec{b} \), the dot product formula simplifies to \( \cos \theta = \vec{a} \cdot \vec{b} \).
This simplification emerges because the magnitudes are both 1, eliminating unnecessary multiplication. Thus, finding the angle \( \theta \) only requires solving for \( \cos \theta \).

• If \( \vec{a} \cdot \vec{b} = \frac{1}{2} \), as in the problem, then \( \cos \theta = \frac{1}{2} \).

The angle whose cosine is \( \frac{1}{2} \) is \( \frac{\pi}{3} \), which corresponds to 60 degrees in more familiar terms. Knowing how to compute this simplifies tasks related to navigation, physics applications, and geometry. The process is key for solving practical questions about angles and positions in a three-dimensional space.