Problem 57 - Master Resource Book in JEE Main Physics

Question

What is the angle between \((\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}})\) 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 \(\cos^{-1}\left(\frac{1}{3}\right)\), so the answer is (d) None of these.

Step 1: Understand the Problem

We need to determine the angle between two vectors. One vector is \((\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}})\) and the other is \(\hat{\mathbf{i}}\). This is done using the dot product formula for vectors.

Step 2: Recall the Dot Product Formula

The dot product \(\vec{A} \cdot \vec{B}\) in terms of vectors \(\vec{A} = \langle a_1, a_2, a_3 \rangle\) and \(\vec{B} = \langle b_1, b_2, b_3 \rangle\) is given by the formula: \(a_1b_1 + a_2b_2 + a_3b_3 = |\vec{A}| |\vec{B}| \cos(\theta)\), where \(\theta\) is the angle between the vectors.

Step 3: Apply the Dot Product Formula

For vectors \(\vec{A} = \langle 1, 2, 2 \rangle\) and \(\vec{B} = \langle 1, 0, 0 \rangle\), the dot product is: \[\vec{A} \cdot \vec{B} = 1\cdot1 + 2\cdot0 + 2\cdot0 = 1.\]

Step 4: Calculate the Magnitudes of Vectors

The magnitude of \(\vec{A}\) is \[|\vec{A}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{9} = 3.\] The magnitude of \(\vec{B}\) is \(|\vec{B}| = \sqrt{1^2 + 0^2 + 0^2} = 1.\)

Step 5: Solve for \(\cos(\theta)\)

Using the dot product equation, \(\vec{A} \cdot \vec{B} = |\vec{A}| |\vec{B}| \cos(\theta)\), we have \(1 = 3\cdot1\cdot\cos(\theta)\). Solving for \(\cos(\theta)\), we get \(\cos(\theta) = \frac{1}{3}\).

Step 6: Find \(\theta\)

The angle \(\theta\) can be found using \(\theta = \cos^{-1}\left(\frac{1}{3}\right)\). Check if it matches the given options. Since \(\pi / 6\approx0.52\) radians and \(\pi / 3\approx1.05\), neither matches \(\cos^{-1}(1/3)\), thus the answer is "None of these."

Key Concepts

Understanding the Angle Between Vectors

The concept of the angle between two vectors is an important aspect of vector algebra. When two vectors are placed tail-to-tail, the angle between them helps in determining how far apart they are directionally. This angle can be found using the dot product of the vectors.
When dealing with the vectors \( \vec{A} = \langle 1, 2, 2 \rangle \) and \( \vec{B} = \langle 1, 0, 0 \rangle \), the interest is in finding the angle \( \theta \) formed between them. In 3D space, this helps in understanding vectors' orientations in relation to each other.
Remember, the smaller the angle, the more directionally aligned the vectors are, while an angle larger than 0 but less than 90 degrees indicates they are less aligned.

Calculating Vector Magnitudes

To calculate the magnitude of a vector in 3D space, we use the formula \( |\vec{A}| = \sqrt{a_1^2 + a_2^2 + a_3^2} \). Magnitude essentially gives the length or the "size" of the vector.
For example, the magnitude of \( \vec{A} = \langle 1, 2, 2 \rangle \) is calculated as follows:

• First, square each component of the vector: \( 1^2 = 1 \), \( 2^2 = 4 \), and another \( 2^2 = 4 \).
• Next, sum these squares: \( 1 + 4 + 4 = 9 \).
• Finally, take the square root of the sum: \( \sqrt{9} = 3 \).

This results in \( |\vec{A}| = 3 \). For the vector \( \vec{B} = \langle 1, 0, 0 \rangle \), the calculation is straightforward: \( |\vec{B}| = 1 \). These magnitudes are crucial for finding the angle between them using the dot product formula.

Using the Cosine of Angle Formula

The cosine of the angle formula is a key tool when finding the angle between two vectors. It states that \( \vec{A} \cdot \vec{B} = |\vec{A}| |\vec{B}| \cos(\theta) \), where \( \theta \) is the angle between the vectors.
This formula connects the dot product and the magnitudes to the angle, allowing us to solve for \( \theta \).
Given \( \vec{A} \cdot \vec{B} = 1 \), \( |\vec{A}| = 3 \), and \( |\vec{B}| = 1 \), substitute these into the formula:

• Calculate \( \cos(\theta) = \frac{1}{3 \times 1} = \frac{1}{3} \).
• To find \( \theta \), use the inverse cosine function: \( \theta = \cos^{-1}\left(\frac{1}{3}\right) \).

This process helps in determining the angular relationship between two vectors, enhancing our understanding of their spatial orientation. In this specific problem, the calculated angle does not match the provided options exactly, indicating the need for precise calculations.