Problem 33 - Master Resource Book in JEE Main Physics

Question

Assertion Angle between \(\hat{\mathbf{i}}+\hat{\mathbf{j}}\) and \(\hat{\mathbf{i}}\) is \(45^{\circ}\) Reason \(\hat{\mathbf{i}}+\hat{\mathbf{j}}\) is equally include to both \(\hat{\mathbf{i}}\) and \(\hat{\mathbf{j}}\) and the angle between \(\hat{i}\) and \(\hat{j}\) is \(90^{\circ}\).

Step-by-Step Solution

Answer

Yes, the assertion is true and supported by the given reason.

Step 1: Understand the Vectors

We have two vectors to consider: \( \hat{\mathbf{i}} + \hat{\mathbf{j}} \) and \( \hat{\mathbf{i}} \). The vector \( \hat{\mathbf{i}} + \hat{\mathbf{j}} \) is formed by adding the unit vectors in the x and y directions.

Step 2: Calculate the Dot Product

The dot product of two vectors \( \mathbf{a} \) and \( \mathbf{b} \) can be calculated using \( \mathbf{a} \cdot \mathbf{b} = a_x b_x + a_y b_y + a_z b_z \). For our vectors \( \hat{\mathbf{i}} = (1, 0) \) and \( \hat{\mathbf{i}} + \hat{\mathbf{j}} = (1, 1) \), the dot product is: \( 1 \times 1 + 0 \times 1 = 1 \).

Step 3: Find the Magnitude of the Vectors

The magnitude of a vector \( \mathbf{a} = (a_x, a_y) \) is given by \( \| \mathbf{a} \| = \sqrt{a_x^2 + a_y^2} \). For \( \hat{\mathbf{i}} \), the magnitude is \( \sqrt{1^2 + 0^2} = 1 \). For \( \hat{\mathbf{i}} + \hat{\mathbf{j}} \), the magnitude is \( \sqrt{1^2 + 1^2} = \sqrt{2} \).

Step 4: Calculate the Angle Using the Dot Product Formula

The angle \( \theta \) between two vectors \( \mathbf{a} \) and \( \mathbf{b} \) can be found using \( \cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\| \|\mathbf{b}\|} \). Here, \( \cos \theta = \frac{1}{1 \cdot \sqrt{2}} = \frac{1}{\sqrt{2}} \). Therefore, \( \theta = 45^{\circ} \).

Step 5: Verify the Reason

The assertion that the angle between \( \hat{\mathbf{i}} + \hat{\mathbf{j}} \) and \( \hat{\mathbf{i}} \) is \( 45^{\circ} \) holds true. The reason provided that \( \hat{\mathbf{i}} + \hat{\mathbf{j}} \) is equally inclined to both \( \hat{\mathbf{i}} \) and \( \hat{\mathbf{j}} \), and since the angle between \( \hat{\mathbf{i}} \) and \( \hat{\mathbf{j}} \) is \( 90^{\circ} \), this implies equal inclination hence supporting the assertion.

Key Concepts

Dot Product

The dot product is a fundamental operation in vector analysis. It allows us to determine the extent to which two vectors point in the same direction. The dot product of two vectors \( \mathbf{a} = (a_x, a_y, a_z) \) and \( \mathbf{b} = (b_x, b_y, b_z) \) is computed as:

• \( \mathbf{a} \cdot \mathbf{b} = a_x b_x + a_y b_y + a_z b_z \)

In our example, we have two vectors:
\( \hat{\mathbf{i}} = (1, 0) \) and \( \hat{\mathbf{i}} + \hat{\mathbf{j}} = (1, 1) \).
To find the dot product, we perform the following calculation:

• \( 1 \times 1 + 0 \times 1 = 1 \)

This result gives us valuable information about the relationship between the vectors,
and it is a crucial step in finding the angle between them.

Vector Magnitude

Vector magnitude, also referred to as the length or norm of a vector,
gives us an idea of how long a vector is. To find the magnitude of a vector \( \mathbf{a} = (a_x, a_y) \),
we use the formula:

• \( \| \mathbf{a} \| = \sqrt{a_x^2 + a_y^2} \)

Let's apply this to our vectors:

• For \( \hat{\mathbf{i}} = (1, 0) \), the magnitude is \( \sqrt{1^2 + 0^2} = 1 \).
• For \( \hat{\mathbf{i}} + \hat{\mathbf{j}} = (1, 1) \), the magnitude is \( \sqrt{1^2 + 1^2} = \sqrt{2} \).

Understanding the magnitude helps us to normalize vectors,
which is necessary for calculating the angle between them.

Angle Between Vectors

The angle between vectors tells us about their relative direction.
To find this angle, we make use of the formula involving the dot product:

• \( \cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\| \|\mathbf{b}\|} \)

Here, \( \theta \) is the angle between vectors \( \mathbf{a} \) and \( \mathbf{b} \).
By rearranging the formula, we can solve for \( \theta \) when the other values are known.
Let's consider our vectors \( \hat{\mathbf{i}} = (1, 0) \) and \( \hat{\mathbf{i}} + \hat{\mathbf{j}} = (1, 1) \).
Previously, we found their dot product to be 1 and their magnitudes to be 1 and \( \sqrt{2} \) respectively.
Applying these values into our formula, we get:

• \( \cos \theta = \frac{1}{1 \times \sqrt{2}} = \frac{1}{\sqrt{2}} \)

Taking the inverse cosine results in \( \theta = 45^{\circ} \),
showing that the two vectors form a 45-degree angle between them.