Problem 65 - Master Resource Book in JEE Main Physics

Question

If the magnitudes of scalar and vector products of two vectors are 6 and \(6 \sqrt{3}\) respectively, then the angle between two vectors is (a) \(15^{\circ}\) (b) \(30^{\circ}\) (c) \(60^{\circ}\) (d) \(75^{\circ}\)

Step-by-Step Solution

Answer

The angle between the vectors is \( 60^{\circ} \).

Step 1: Understand the Problem

We have two vectors and we need to find out the angle between them given that the magnitudes of their scalar product (dot product) and vector product (cross product) are 6 and \( 6 \sqrt{3} \) respectively.

Step 2: Identify Formulas to Use

To solve this, we use the formulas for dot product and cross product:- Dot product: \( \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos(\theta) = 6 \)- Cross product magnitude: \( |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin(\theta) = 6 \sqrt{3} \)

Step 3: Express Trigonometric Values

Divide the cross product magnitude by the dot product:\[\frac{|\vec{a} \times \vec{b}|}{\vec{a} \cdot \vec{b}} = \frac{|\vec{a}| |\vec{b}| \sin(\theta)}{|\vec{a}| |\vec{b}| \cos(\theta)} = \sqrt{3}\]This simplifies to \( \tan(\theta) = \sqrt{3} \).

Step 4: Solve for the Angle

Since \( \tan(\theta) = \sqrt{3} \), we look for the angle where the tangent equals \( \sqrt{3} \). This angle is known to be \( 60^{\circ} \).

Step 5: Verify the Solution

Check that at \( 60^{\circ} \), the products match the given magnitudes. Substitute \( \theta = 60^{\circ} \) into both product formulas:- \( \cos(60^{\circ}) = \frac{1}{2}, \sin(60^{\circ}) = \frac{\sqrt{3}}{2} \)- Scalar product is consistent: \( 6 = |\vec{a}| |\vec{b}| \cdot \frac{1}{2} \)- Cross product is consistent: \( 6\sqrt{3} = |\vec{a}| |\vec{b}| \cdot \frac{\sqrt{3}}{2} \)

Key Concepts

Understanding Scalar and Vector Products

To understand vector mathematics, it is essential to grasp the concepts of scalar and vector products. A scalar product, also known as the dot product, multiplies two vectors to give a scalar, a single-number result. This is calculated using the formula:

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

where \( |\vec{a}| \) and \( |\vec{b}| \) are the magnitudes of vectors \( \vec{a} \) and \( \vec{b} \), and \( \theta \) is the angle between them. The cosine function signifies the relationship between the direction of the vectors.

On the other hand, a vector product, or cross product, results in another vector perpendicular to the plane containing the original two vectors. The magnitude of this vector is given by:

• \( |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin(\theta) \)

The sine function emphasizes that this product varies depending on the angle. These products provide different types of information about the vectors involved, crucial for understanding how vectors interact in space.

Finding the Angle Between Vectors

Determining the angle between vectors involves understanding how scalar and vector products contribute to the spatial relationship. Given the situation where the products are known, we can use the ratio of the cross product magnitude to the dot product, leading to:

• \( \frac{|\vec{a} \times \vec{b}|}{\vec{a} \cdot \vec{b}} = \frac{|\vec{a}| |\vec{b}| \sin(\theta)}{|\vec{a}| |\vec{b}| \cos(\theta)} = \tan(\theta) \)

This simple manipulation shows that the tangent of the angle determines the angle itself when the magnitudes of the products are known. In our specific problem, solving \( \tan(\theta) = \sqrt{3} \) leads us to the angle \( \theta = 60^{\circ} \).
The evaluation of tangent allows us to rely on trigonometric identities, ensuring the angle is accurate regarding the vector orientations.

Exploring Trigonometric Values

Trigonometric functions such as sine, cosine, and tangent are essential in vector analysis. They reveal how angles and their measures relate to vector magnitudes.

• Cosine \( \cos(\theta) \): Determines the ratio of the adjacent side to the hypotenuse in a right triangle.
• Sine \( \sin(\theta) \): Relates the opposite side of a triangle to the hypotenuse.
• Tangent \( \tan(\theta) \): Connects the opposite side to the adjacent side.

In the exercise, recognizing \( \tan(\theta) = \sqrt{3} \) and linking it to the known trigonometric value for \( 60^{\circ} \) is key. Given that \( \cos(60^{\circ}) = \frac{1}{2} \) and \( \sin(60^{\circ}) = \frac{\sqrt{3}}{2} \), these established values allow solutions to be checked and verified through known calculations. Understanding these basics supports complex problem-solving in physics and beyond, offering clarity in abstract vector concepts.