Problem 34 - Master Resource Book in JEE Main Physics

Question

Assertion The vector \(\frac{1}{\sqrt{3}} \hat{\mathbf{i}}+\frac{1}{\sqrt{3}} \hat{\mathbf{j}}+\frac{1}{\sqrt{3}} \hat{\mathbf{k}}\) is a unit vector. Reason Unit vector is one which has unit magnitude and a given direction.

Step-by-Step Solution

Answer

The vector is a unit vector as its magnitude is 1.

Step 1: Identify the Vector

We are given the vector \( \mathbf{v} = \left( \frac{1}{\sqrt{3}} \right) \hat{\mathbf{i}} + \left( \frac{1}{\sqrt{3}} \right) \hat{\mathbf{j}} + \left( \frac{1}{\sqrt{3}} \right) \hat{\mathbf{k}} \). Our task is to determine whether this vector is indeed a unit vector.

Step 2: Recollect the Definition of a Unit Vector

A unit vector is a vector whose magnitude is equal to 1. It is typically denoted as \( \hat{\mathbf{v}} \) and maintains the direction of the original vector.

Step 3: Compute the Magnitude of the Vector

The magnitude of a vector \( \mathbf{v} = a \hat{\mathbf{i}} + b \hat{\mathbf{j}} + c \hat{\mathbf{k}} \) is calculated using the formula: \( |\mathbf{v}| = \sqrt{a^2 + b^2 + c^2} \). Substitute the coefficients of \( \mathbf{v} \): \[ |\mathbf{v}| = \sqrt{\left( \frac{1}{\sqrt{3}} \right)^2 + \left( \frac{1}{\sqrt{3}} \right)^2 + \left( \frac{1}{\sqrt{3}} \right)^2} \].

Step 4: Simplify and Verify the Magnitude

Simplify the expression: \[ |\mathbf{v}| = \sqrt{\frac{1}{3} + \frac{1}{3} + \frac{1}{3}} = \sqrt{1} = 1 \]. Since the magnitude is 1, the vector is indeed a unit vector.

Key Concepts

Magnitude of a Vector

The magnitude of a vector is a measure of its length in space. Think of it as the size or extent of the vector. It's calculated using a formula based on the components of the vector. If you have a vector written in terms of its components like \[ \mathbf{v} = a \hat{\mathbf{i}} + b \hat{\mathbf{j}} + c \hat{\mathbf{k}} \],its magnitude \(|\mathbf{v}|\) is determined by finding the square root of the sum of the squares of its components:\[ |\mathbf{v}| = \sqrt{a^2 + b^2 + c^2} \].
Each of the components \(a\), \(b\), and \(c\) is squared, summed up, and then we take the square root of that total sum.

• This square root operation ensures that the magnitude remains a non-negative number.
• The magnitude gives the vector a size but no direction.

To illustrate, consider a vector \( \left( \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \right) \). To find its magnitude, we'd compute:\[ |\mathbf{v}| = \sqrt{\left( \frac{1}{\sqrt{3}} \right)^2 + \left( \frac{1}{\sqrt{3}} \right)^2 + \left( \frac{1}{\sqrt{3}} \right)^2} = 1 \].
This confirms it has a magnitude of 1, illustrating it's a unit vector.

Vector Components

Vector components are projections of a vector along the axes of the coordinate system.Each vector in 3D space can be represented as a sum of its components along the \(x\), \(y\), and \(z\) axes.
Suppose vector \( \mathbf{v} \) is given by:\[ \mathbf{v} = a \hat{\mathbf{i}} + b \hat{\mathbf{j}} + c \hat{\mathbf{k}} \]

• \(a\) is the component along the \(x\)-axis and corresponds to the impact in the direction \(\hat{\mathbf{i}}\).
• \(b\) is the component along the \(y\)-axis related to the direction \(\hat{\mathbf{j}}\).
• \(c\) is the component along the \(z\)-axis in the direction \(\hat{\mathbf{k}}\).

These components are like breaking the vector down into smaller, easy-to-handle parts that align perfectly with our coordinate grid.
In the case of our vector \( \left( \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \right) \), we have equal components in all directions, making it quite symmetric and balanced in 3D space.

Definition of a Unit Vector

A unit vector is a special type of vector. It makes things simpler by having a magnitude equal to 1. A unit vector maintains the direction of the original vector but with a standardized length of 1.
The representation of a unit vector is usually symbolized using a hat, such as \( \hat{\mathbf{v}} \). The process of finding a unit vector, which points in the same direction as a given vector, is called normalization.

• You take the original vector and divide each component by the vector’s magnitude.
• This scales the vector down (or up) to a length of 1 while keeping its direction.

For example, if vector \( \mathbf{v} = \left( a, b, c \right) \), to find its unit vector \( \hat{\mathbf{v}} \),you would calculate:\[ \hat{\mathbf{v}} = \left( \frac{a}{|\mathbf{v}|}, \frac{b}{|\mathbf{v}|}, \frac{c}{|\mathbf{v}|} \right) \].In contexts where direction is important, unit vectors play a critical role because they provide the orientation without any scaling concerns.