Problem 53 - Master Resource Book in JEE Main Physics

Question

If, \(\mathbf{A}=2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}\) and \(\mathbf{B}=4 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}\), then angle between \(\mathbf{A}\) and \(\mathbf{B}\) is (a) \(\sin ^{-1}\left(\frac{25}{29}\right)\) (b) \(\sin ^{-1}\left(\frac{29}{25}\right)\) (c) \(\cos ^{-1}\left(\frac{25}{29}\right)\) (d) \(\cos ^{-1}\left(\frac{29}{25}\right)\)

Step-by-Step Solution

Answer

The angle is \( \cos^{-1} \left( \frac{25}{29} \right) \); option (c).

Step 1: Compute dot product of A and B

First, we need to find the dot product \( \mathbf{A} \cdot \mathbf{B} \). The formula for the dot product of two vectors \( \mathbf{A} = a_1 \hat{\mathbf{i}} + a_2 \hat{\mathbf{j}} + a_3 \hat{\mathbf{k}} \) and \( \mathbf{B} = b_1 \hat{\mathbf{i}} + b_2 \hat{\mathbf{j}} + b_3 \hat{\mathbf{k}} \) is \( a_1b_1 + a_2b_2 + a_3b_3 \). For the vectors \( \mathbf{A} = 2 \hat{\mathbf{i}} + 3 \hat{\mathbf{j}} + 4 \hat{\mathbf{k}} \) and \( \mathbf{B} = 4 \hat{\mathbf{i}} + 3 \hat{\mathbf{j}} + 2 \hat{\mathbf{k}} \), this becomes \( 2\times4 + 3\times3 + 4\times2 = 8 + 9 + 8 = 25 \). Thus, \( \mathbf{A} \cdot \mathbf{B} = 25 \).

Step 2: Calculate magnitudes of A and B

Next, compute the magnitudes of vectors \( \mathbf{A} \) and \( \mathbf{B} \). The magnitude \( \|\mathbf{A}\| \) is given by \( \sqrt{a_1^2 + a_2^2 + a_3^2} \), so \( \|\mathbf{A}\| = \sqrt{2^2 + 3^2 + 4^2} = \sqrt{4 + 9 + 16} = \sqrt{29} \). Similarly, \( \|\mathbf{B}\| = \sqrt{4^2 + 3^2 + 2^2} = \sqrt{16 + 9 + 4} = \sqrt{29} \).

Step 3: Use cosine formula for angle

The cosine of the angle \( \theta \) between two vectors \( \mathbf{A} \) and \( \mathbf{B} \) is given by the formula \( \cos \theta = \frac{\mathbf{A} \cdot \mathbf{B}}{\|\mathbf{A}\| \|\mathbf{B}\|} \). We have \( \mathbf{A} \cdot \mathbf{B} = 25 \) and both magnitudes are \( \sqrt{29} \). So \( \cos \theta = \frac{25}{\sqrt{29} \times \sqrt{29}} = \frac{25}{29} \).

Step 4: Choose correct inverse function

Since the problem asks for the angle and provides choices in inverse trigonometric forms, use \( \cos^{-1}\) for \(\theta\). Thus, the angle \( \theta = \cos^{-1} \left( \frac{25}{29} \right) \).

Key Concepts

dot product

The dot product is a fundamental operation in vector algebra. It allows us to multiply two vectors and results in a scalar (a single number). Given two vectors, \( \mathbf{A} = a_1 \hat{\mathbf{i}} + a_2 \hat{\mathbf{j}} + a_3 \hat{\mathbf{k}} \) and \( \mathbf{B} = b_1 \hat{\mathbf{i}} + b_2 \hat{\mathbf{j}} + b_3 \hat{\mathbf{k}} \), the dot product is calculated using:

• \( \mathbf{A} \cdot \mathbf{B} = a_1b_1 + a_2b_2 + a_3b_3 \)

This operation combines corresponding components of the vectors. For the vectors given, \( \mathbf{A} = 2 \hat{\mathbf{i}} + 3 \hat{\mathbf{j}} + 4 \hat{\mathbf{k}} \) and \( \mathbf{B} = 4 \hat{\mathbf{i}} + 3 \hat{\mathbf{j}} + 2 \hat{\mathbf{k}} \), the dot product becomes \( 2\times4 + 3\times3 + 4\times2 = 25 \).

This result is essential because it is used to measure the degree to which two vectors are parallel. If the dot product is positive, it indicates that the vectors are pointing in generally the same direction. A negative result would imply opposite directions, and if the dot product is zero, the vectors are perpendicular.

magnitude of vectors

The magnitude of a vector is akin to its length in space. It conveys how far the vector's direction extends from the origin, regardless of its orientation. Calculating the magnitude, or lengths of vectors \( \mathbf{A} \) and \( \mathbf{B} \), is crucial for various calculations, including finding angles between vectors.

• The formula for the magnitude of a vector \( \mathbf{V} = v_1 \hat{\mathbf{i}} + v_2 \hat{\mathbf{j}} + v_3 \hat{\mathbf{k}} \) is \( \|\mathbf{V}\| = \sqrt{v_1^2 + v_2^2 + v_3^2} \).

Applying this to \( \mathbf{A} = 2 \hat{\mathbf{i}} + 3 \hat{\mathbf{j}} + 4 \hat{\mathbf{k}} \), we get \( \|\mathbf{A}\| = \sqrt{2^2 + 3^2 + 4^2} = \sqrt{29} \). Similarly, for \( \mathbf{B} = 4 \hat{\mathbf{i}} + 3 \hat{\mathbf{j}} + 2 \hat{\mathbf{k}} \), its magnitude is \( \|\mathbf{B}\| = \sqrt{29} \).

The magnitude quantitative tells us that both vectors are equally 'long' in the vectorial sense, although they might point in different directions. This understanding of magnitude is foundational for calculating the angles between vectors using the cosine formula.

cosine formula for angle

Calculating the angle between vectors is a common application of the dot product and the magnitude of vectors. The cosine formula helps find the angle \( \theta \) between two vectors \( \mathbf{A} \) and \( \mathbf{B} \). Based on vector knowledge, the formula is written as:

• \( \cos \theta = \frac{\mathbf{A} \cdot \mathbf{B}}{\|\mathbf{A}\| \|\mathbf{B}\|} \)

Given that we already computed the dot product \( \mathbf{A} \cdot \mathbf{B} = 25 \) and know the magnitudes \( \|\mathbf{A}\| = \sqrt{29} \) and \( \|\mathbf{B}\| = \sqrt{29} \), we can substitute these into the formula to get \( \cos \theta = \frac{25}{\sqrt{29} \times \sqrt{29}} = \frac{25}{29} \).

Finally, using an inverse cosine function, \( \theta = \cos^{-1} \left( \frac{25}{29} \right) \), we find the angle between the vectors. This angle tells us how divergent the vectors are in space. With this knowledge, one can determine the extent of parallelism or orthogonality between the vectors, which is critical in various fields, such as physics and engineering.