Problem 53 - Fundamentals of Physics

Question

A vector \(\vec{a}\) of magnitude 10 units and another vector \(\vec{b}\) of magnitude \(6.0\) units differ in directions by \(60^{\circ} .\) Find (a) the scalar product of the two vectors and (b) the magnitude of the vector product \(\vec{a} \times \vec{b}\)

Step-by-Step Solution

Answer

(a) 30, (b) 30√3

Step 1: Understanding Scalar Product Formula

The scalar product (or dot product) of two vectors \(\vec{a}\) and \(\vec{b}\) is given by the formula \(\vec{a} \cdot \vec{b} = |\vec{a}| \, |\vec{b}| \, \cos(\theta)\), where \(\theta\) is the angle between the vectors. We are given that \(|\vec{a}| = 10\) units, \(|\vec{b}| = 6\) units, and \(\theta = 60^\circ\).

Step 2: Calculating the Scalar Product

Substitute the known values into the formula: \(\vec{a} \cdot \vec{b} = 10 \times 6 \times \cos(60^\circ)\). Since \(\cos(60^\circ) = \frac{1}{2}\), the calculation becomes \(\vec{a} \cdot \vec{b} = 10 \times 6 \times \frac{1}{2} = 30\).

Step 3: Understanding Vector Product Magnitude Formula

The magnitude of the vector product (or cross product) \(\vec{a} \times \vec{b}\) is given by \(|\vec{a} \times \vec{b}| = |\vec{a}| \, |\vec{b}| \, \sin(\theta)\). We will use the same values for \(\vec{a}\), \(\vec{b}\), and \(\theta\) as before.

Step 4: Calculating the Magnitude of the Vector Product

Substitute the known values into the formula: \(|\vec{a} \times \vec{b}| = 10 \times 6 \times \sin(60^\circ)\). Knowing that \(\sin(60^\circ) = \frac{\sqrt{3}}{2}\), the calculation becomes \(|\vec{a} \times \vec{b}| = 10 \times 6 \times \frac{\sqrt{3}}{2} = 30\sqrt{3}\).

Key Concepts

Scalar Product

The scalar product, also known as the dot product, is a way to multiply two vectors, resulting in a scalar (a single number). It is calculated using the formula:

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

Here, \(|\vec{a}|\) and \(|\vec{b}|\) are the magnitudes (lengths) of the vectors \(\vec{a}\) and \(\vec{b}\), and \(\theta\) is the angle between them.

In our exercise, we are given that

• \(|\vec{a}| = 10\) units
• \(|\vec{b}| = 6\) units
• \(\theta = 60^\circ\)

Using these values, we substitute into the formula:

\( \vec{a} \cdot \vec{b} = 10 \times 6 \times \cos(60^\circ) \)

Knowing that \(\cos(60^\circ) = \frac{1}{2}\), the calculation simplifies to:

• \( \vec{a} \cdot \vec{b} = 10 \times 6 \times \frac{1}{2} = 30 \) units

This result shows how the scalar product can effectively represent how much one vector points in the direction of another.

Vector Product

The vector product, or cross product, of two vectors results not in a scalar but in another vector, which is perpendicular to the plane containing the original vectors. To find the magnitude of this vector, we use the formula:

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

In this formula, \(|\vec{a}|\) and \(|\vec{b}|\) are the magnitudes of the vectors, and \(\theta\) is the angle between them. The sine function reflects how much of one vector is perpendicular to the other.

For the given problem:

• \(|\vec{a}| = 10\) units
• \(|\vec{b}| = 6\) units
• \(\theta = 60^\circ\)

Substituting these into our formula, we get:

\( |\vec{a} \times \vec{b}| = 10 \times 6 \times \sin(60^\circ) \)

Since \(\sin(60^\circ) = \frac{\sqrt{3}}{2}\), we find:

\( |\vec{a} \times \vec{b}| = 10 \times 6 \times \frac{\sqrt{3}}{2} = 30\sqrt{3} \)

This result gives us the magnitude of the vector that represents the area spanned by the two vectors when they are placed tail-to-tail.

Magnitude of Vectors

To fully understand vector operations, we must comprehend what the magnitude of a vector signifies. The magnitude is simply the length or size of the vector, usually measured in units.

For a vector \(\vec{a}\) with components \(a_x\), \(a_y\), and \(a_z\) in three-dimensional space, the magnitude is determined by:

• \( |\vec{a}| = \sqrt{a_x^2 + a_y^2 + a_z^2} \)

Thus, the magnitude offers a straightforward way to determine how long or big the vector is without considering its direction.

In our specific exercise, vector magnitudes were provided directly as \(|\vec{a}| = 10\) units and \(|\vec{b}| = 6\) units. These values simplify the process of calculating the scalar and vector products, as they provide a direct measure of each vector's size. The magnitude is a key building block of understanding more complex vector operations since it translates a vector, which has both direction and size, into a mere scalar (size or length). This approach allows for easier comparisons and calculations involving vectors.