Problem 38 - Master Resource Book in JEE Main Physics

Question

What is the unit vector along \(\hat{\mathbf{i}}+\hat{\mathbf{j}}\) ? (a) \(\frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}}{\sqrt{2}}\) (b) \(\sqrt{2}(\hat{i}+\hat{j})\) (c) \(\hat{\mathrm{i}}+\hat{\mathrm{j}}\) (d) \(\hat{\mathrm{k}}\)

Step-by-Step Solution

Answer

The unit vector is option (a) \(\frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}}{\sqrt{2}}\).

Step 1: Understand the Problem

We need to find the unit vector along the vector \(\hat{\mathbf{i}} + \hat{\mathbf{j}}\). A unit vector is a vector with a magnitude (or length) of 1 in the direction of a given vector.

Step 2: Calculate the Magnitude of the Vector

Compute the magnitude of the vector \(\hat{\mathbf{i}} + \hat{\mathbf{j}}\). Use the formula for the magnitude of a vector \(\mathbf{v} = a\hat{\mathbf{i}} + b\hat{\mathbf{j}}\), which is \(\sqrt{a^2 + b^2}\). In this case, each component is 1: \(\sqrt{1^2 + 1^2} = \sqrt{2}\).

Step 3: Find the Unit Vector

To find the unit vector, divide each component of the vector by its magnitude. The original vector is \(\hat{\mathbf{i}} + \hat{\mathbf{j}}\) and its magnitude is \(\sqrt{2}\). Thus, the unit vector is \(\frac{1}{\sqrt{2}}\hat{\mathbf{i}} + \frac{1}{\sqrt{2}}\hat{\mathbf{j}}\), or equivalently \(\frac{\hat{\mathbf{i}} + \hat{\mathbf{j}}}{\sqrt{2}}\).

Step 4: Choose the Correct Option

Compare the calculated unit vector \(\frac{\hat{\mathbf{i}} + \hat{\mathbf{j}}}{\sqrt{2}}\) to the given options. The correct option is (a) \(\frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}}{\sqrt{2}}\).

Key Concepts

Vector Magnitude

Magnitude is a measure of a vector's length or size. It tells us how long or big a vector is. To calculate the magnitude of a vector, we use the Pythagorean theorem, which is a well-known method in mathematics. For a two-dimensional vector, like \( \mathbf{v} = a\hat{\mathbf{i}} + b\hat{\mathbf{j}} \), the magnitude is given by the formula:

• \( \sqrt{a^2 + b^2} \)

This applies the theorem to find a 'hypotenuse', or the diagonal line in a right-angled triangle formed by the vector components.
In our specific case, the vector is \( \hat{\mathbf{i}} + \hat{\mathbf{j}} \). Each of these components is equal to one:

• \( \hat{\mathbf{i}} = 1 \)
• \( \hat{\mathbf{j}} = 1 \)

So, the magnitude is calculated as \( \sqrt{1^2 + 1^2} = \sqrt{2} \). The magnitude signifies the distance from the origin to the vector's tip in geometric terms.

Direction of a Vector

The direction of a vector tells us where the vector is pointing. It is an essential aspect of understanding vectors alongside magnitude. The direction is determined by the angle the vector makes with a predefined line, usually the positive x-axis.
For a vector like \( \hat{\mathbf{i}} + \hat{\mathbf{j}} \), it points equally in the x and y directions, forming a diagonal line at a 45-degree angle in the positive quadrant.

• Its ratio \( \frac{y}{x} \) here is 1 since both components are equal.
• This means the vector is at \( 45^\circ \) to the positive x-axis.

Visualizing this can help you see direction as the 'orientation' or 'aim' of a vector. Knowing the direction makes it easier to understand vector components and resultant vectors in physics and engineering problems.

Vector Components

Vectors are often expressed in terms of their components, which are projections along the axes of a coordinate system. For instance, in a 2D plane, any vector can be broken down into its x (horizontal) and y (vertical) components. This breakdown helps us understand the vector's influence along different directions.
A vector \( \mathbf{v} = a\hat{\mathbf{i}} + b\hat{\mathbf{j}} \) has components:

• The x-component: \( a\hat{\mathbf{i}} \)
• The y-component: \( b\hat{\mathbf{j}} \)

In the case of \( \hat{\mathbf{i}} + \hat{\mathbf{j}} \):

• The x-component is \( 1\hat{\mathbf{i}} \)
• The y-component is \( 1\hat{\mathbf{j}} \)

Understanding these components allows us to compute the resulting vector if several vectors are combined. It also makes the problems of finding velocity, acceleration, and forces in specific directions in physics and engineering more tractable.