Problem 66 - Master Resource Book in JEE Main Physics

Question

What is the angle between \(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}\) and \(\hat{\mathbf{i}}\) ? (a) \(0^{\circ}\) (b) \(\pi / 6\) (c) \(\pi / 3\) (d) None of these

Step-by-Step Solution

Answer

The angle is \( \pi/6 \) radians.

Step 1: Understand the Vectors

The vectors provided are \( \hat{\mathbf{i}} + \hat{\mathbf{j}} + \hat{\mathbf{k}} \) and \( \hat{\mathbf{i}} \). These are expressed in terms of components along the x, y, and z axes.

Step 2: Calculate the Dot Product

Use the formula for the dot product: \( \mathbf{a} \cdot \mathbf{b} = a_xb_x + a_yb_y + a_zb_z \). Here, \( \mathbf{a} = \hat{\mathbf{i}} + \hat{\mathbf{j}} + \hat{\mathbf{k}} \) and \( \mathbf{b} = \hat{\mathbf{i}} \). The dot product is \( (1)(1) + (1)(0) + (1)(0) = 1 \).

Step 3: Find the Magnitudes of the Vectors

The magnitude of \( \hat{\mathbf{i}} + \hat{\mathbf{j}} + \hat{\mathbf{k}} \) is \( \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3} \). The magnitude of \( \hat{\mathbf{i}} \) is \( \sqrt{1^2} = 1 \).

Step 4: Use the Dot Product to Find the Cosine of the Angle

The cosine of the angle \( \theta \) between two vectors is given by \( \cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\| \|\mathbf{b}\|} \). Substituting the values, we get \( \cos \theta = \frac{1}{\sqrt{3} \times 1} = \frac{1}{\sqrt{3}} \).

Step 5: Find the Angle from the Cosine Value

The angle whose cosine is \( \frac{1}{\sqrt{3}} \) is \( \pi/6 \) radians. This is obtained by using the inverse cosine function, \( \theta = \cos^{-1}\left(\frac{1}{\sqrt{3}}\right) \).

Key Concepts

Dot Product

The dot product, also known as the scalar product, is a foundational concept in vector mathematics. It is a way to multiply two vectors to produce a single scalar. The formula for the dot product of two vectors \( \mathbf{a} = (a_x, a_y, a_z) \) and \( \mathbf{b} = (b_x, b_y, b_z) \) is: \( \mathbf{a} \cdot \mathbf{b} = a_xb_x + a_yb_y + a_zb_z \).

This equation tells us that in order to find the dot product, we simply multiply the corresponding components of the vectors and add them together. The result is a single number, which can be used to determine properties such as the angle between two vectors. In the case of our example, the vector \( \hat{\mathbf{i}} + \hat{\mathbf{j}} + \hat{\mathbf{k}} \) and \( \hat{\mathbf{i}} \) have a dot product of 1. This step is crucial because it bridges vector multiplication with geometric analysis.

• Determine corresponding vector components for multiplication.
• Add the results to get the dot product.

Magnitude of a Vector

The magnitude of a vector is a measure of its length in space. It is equivalent to the "distance" the vector represents from the origin. The formula for finding the magnitude of a given vector \( \mathbf{v} = (v_x, v_y, v_z) \) is \( \|\mathbf{v}\| = \sqrt{v_x^2 + v_y^2 + v_z^2} \).

This means to find how "long" the vector is, you square each of its components, add them together, and then take the square root of that summation. In our example, the magnitude of \( \hat{\mathbf{i}} + \hat{\mathbf{j}} + \hat{\mathbf{k}} \) is \( \sqrt{3} \), while the magnitude of \( \hat{\mathbf{i}} \) is 1.

• Square each component of the vector.
• Add those squared values.
• Take the square root of the sum to find the magnitude.

Trigonometry in Vectors

Trigonometry plays a key role in understanding and calculating angles between vectors. Using the dot product and magnitudes, we can find the angle between two vectors using the cosine formula: \( \cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\| \|\mathbf{b}\|} \).

This formula is derived from the dot product definition and captures how the angle relates to vector magnitudes. In the given problem, \( \cos \theta = \frac{1}{\sqrt{3}} \), and subsequently, the angle \( \theta \) is found by taking the inverse cosine \( \cos^{-1} \). The result is \( \pi/6 \) radians, highlighting how trigonometry helps us transition from vector calculations to actual angle determination.

• Use the dot product and magnitudes to set up \( \cos \theta \).
• Calculate \( \theta \) using the inverse cosine function.
• Understand how this connects the geometry and algebra of vectors.