Two vector `vecA and vecB` of magnitude 50 N and 100N are inclined at an angle of `60^(@)`. Calculate the magnitude of cross product and dot product of these two vectors.

Answer

To solve the problem of calculating the magnitude of the cross product and dot product of two vectors \(\vec{A}\) and \(\vec{B}\) with magnitudes of 50 N and 100 N, respectively, and an angle of \(60^\circ\) between them, we can follow these steps:

### Step 1: Calculate the Dot Product

The formula for the dot product of two vectors \(\vec{A}\) and \(\vec{B}\) is given by:

\[ \vec{A} \cdot \vec{B} = |\vec{A}| |\vec{B}| \cos(\theta) \]

Where:
- \(|\vec{A}| = 50 \, \text{N}\)
- \(|\vec{B}| = 100 \, \text{N}\)
- \(\theta = 60^\circ\)

Substituting the values into the formula:

\[ \vec{A} \cdot \vec{B} = 50 \times 100 \times \cos(60^\circ) \]

Since \(\cos(60^\circ) = \frac{1}{2}\):

\[ \vec{A} \cdot \vec{B} = 50 \times 100 \times \frac{1}{2} \]

Calculating this gives:

\[ \vec{A} \cdot \vec{B} = 2500 \, \text{N} \]

### Step 2: Calculate the Cross Product

The formula for the cross product of two vectors \(\vec{A}\) and \(\vec{B}\) is given by:

\[ \vec{A} \times \vec{B} = |\vec{A}| |\vec{B}| \sin(\theta) \]

Using the same magnitudes and angle:

\[ \vec{A} \times \vec{B} = 50 \times 100 \times \sin(60^\circ) \]

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

\[ \vec{A} \times \vec{B} = 50 \times 100 \times \frac{\sqrt{3}}{2} \]

Calculating this gives:

\[ \vec{A} \times \vec{B} = 2500\sqrt{3} \, \text{N} \]

### Final Answers

- The magnitude of the dot product \(\vec{A} \cdot \vec{B} = 2500 \, \text{N}\)
- The magnitude of the cross product \(\vec{A} \times \vec{B} = 2500\sqrt{3} \, \text{N}\)