Problem 44 - Master Resource Book in JEE Main Physics

Question

The angle subtended by the vector, \(\mathbf{A}=4 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+12 \hat{\mathbf{k}}\) with the \(x\)-axis is (a) \(\sin ^{-1}\left(\frac{3}{13}\right)\) (b) \(\sin ^{-1}\left(\frac{4}{13}\right)\) (c) \(\cos ^{-1}\left(\frac{4}{13}\right)\) (d) \(\cos ^{-1}\left(\frac{3}{13}\right)\)

Step-by-Step Solution

Answer

The correct answer is (c) \(\cos^{-1}\left(\frac{4}{13}\right)\).

Step 1: Understand the Problem

We need to find the angle that the given vector \( \mathbf{A} = 4 \hat{\mathbf{i}} + 3 \hat{\mathbf{j}} + 12 \hat{\mathbf{k}} \) makes with the \( x \)-axis. This angle can be found using the dot product between vector \( \mathbf{A} \) and the unit vector along the \( x \)-axis, \( \hat{\mathbf{i}} \).

Step 2: Calculate the Magnitude of Vector \( \mathbf{A} \)

The magnitude of vector \( \mathbf{A} \) is calculated using the formula for the magnitude of a vector, \[ |\mathbf{A}| = \sqrt{(4)^2 + (3)^2 + (12)^2} = \sqrt{16 + 9 + 144} = \sqrt{169} = 13. \]

Step 3: Determine the Dot Product

The dot product of \( \mathbf{A} \) and \( \hat{\mathbf{i}} \) is \[ \mathbf{A} \cdot \hat{\mathbf{i}} = 4. \] This is because the \( x \)-component of \( \mathbf{A} \) is 4, and the other components are zero when dotting with \( \hat{\mathbf{i}} \).

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

The cosine of the angle \( \theta \) that \( \mathbf{A} \) makes with the \( x \)-axis is given by \[ \cos \theta = \frac{\mathbf{A} \cdot \hat{\mathbf{i}}}{|\mathbf{A}|}. \] Substituting the known values gives \[ \cos \theta = \frac{4}{13}. \]

Step 5: Find the Angle

To find the angle \( \theta \), we use the inverse cosine (arc cosine) function: \[ \theta = \cos^{-1}\left(\frac{4}{13}\right). \]

Step 6: Match to the Given Options

Comparing with the given options, the angle \( \theta = \cos^{-1}\left(\frac{4}{13}\right) \) matches option (c).

Key Concepts

Dot Product

The dot product is a significant concept in vector mathematics. It allows us to compute a single number from two vectors, and it's a crucial tool in determining angles between vectors. The dot product of two vectors, say \( \mathbf{A} = a_1 \hat{\mathbf{i}} + a_2 \hat{\mathbf{j}} + a_3 \hat{\mathbf{k}} \) and \( \mathbf{B} = b_1 \hat{\mathbf{i}} + b_2 \hat{\mathbf{j}} + b_3 \hat{\mathbf{k}} \), is given by the formula:

• \( \mathbf{A} \cdot \mathbf{B} = a_1 b_1 + a_2 b_2 + a_3 b_3 \)

For example, if you want to find the dot product of a vector \( \mathbf{A} = 4 \hat{\mathbf{i}} + 3 \hat{\mathbf{j}} + 12 \hat{\mathbf{k}} \) with the unit vector along the \( x \)-axis \( \hat{\mathbf{i}} \), the formula simplifies because \( \hat{\mathbf{i}} \) has components \( b_1 = 1\) and \( b_2 = 0\), \( b_3 = 0 \). So the dot product becomes:

• \( \mathbf{A} \cdot \hat{\mathbf{i}} = 4 \times 1 + 3 \times 0 + 12 \times 0 = 4 \)

Thus, the dot product helps in isolating the component of vector \( \mathbf{A} \) in the direction of \( x \)-axis.

Vector Magnitude

The magnitude of a vector is a measure of its length in space. It's often referred to as the "length" or "norm" of the vector. To calculate this, you use the formula for a vector \( \mathbf{A} = a_1 \hat{\mathbf{i}} + a_2 \hat{\mathbf{j}} + a_3 \hat{\mathbf{k}} \), which is given by:

• \( |\mathbf{A}| = \sqrt{a_1^2 + a_2^2 + a_3^2} \)

Let's apply this to our vector \( \mathbf{A} = 4 \hat{\mathbf{i}} + 3 \hat{\mathbf{j}} + 12 \hat{\mathbf{k}} \):

• \( |\mathbf{A}| = \sqrt{4^2 + 3^2 + 12^2} = \sqrt{16 + 9 + 144} = \sqrt{169} = 13 \)

This tells us that the vector \( \mathbf{A} \) has a magnitude of 13 units. Understanding the magnitude is essential as it is broadly applied in normalizing vectors and in calculating angles between vectors.

Inverse Cosine Function (Arc Cosine)

To find the angle between vectors, we often use the inverse cosine function, also known as the arc cosine. This function helps us move from a cosine value back to an angle measurement. If you have the cosine of the angle, \( \cos \theta \), and you wish to find \( \theta \), you use:

• \( \theta = \cos^{-1}(\cos \theta) \)

For example, if we have computed that \( \cos \theta = \frac{4}{13} \), then to find \( \theta \), we use the inverse cosine:

• \( \theta = \cos^{-1}\left(\frac{4}{13}\right) \)

This step is crucial because sometimes the angle isn't always intuitive from the cosine value. Especially in scenarios involving angles in three dimensions, using the inverse cosine offers a straightforward means to determine explicit angle measurements. Understanding this function is key in areas of mathematics and physics dealing with rotational motion and direction alignment.