**– $ \overrightarrow A \space \times \overrightarrow B $**

**– Determine the vector product’s direction $ \overrightarrow A \space \times \overrightarrow B$.**

**– Calculate the scalar product when the angle is $ 60 { \circ} $ and the vector magnitude is $ 5 and 4 $.**

**– Calculate the scalar product when the angle is $ 60 { \circ} $ and the vector magnitude is $ 5 \space and \space 5 $.**

The main purpose of this guide is to **find** the **direction and magnitude** of the vector product.

This question uses the concept of** magnitude and direction of vector product**. A vector product has both **magnitude and direction**. Mathematically, the vector product is **represented** as:

\[A \space \times \space B \space = \space ||A || \space || B || \space sin \theta n \]

## Expert Answer

We first have to **find** the **direction and magnitude** of the **vector product**.

a) \[A \space \times \space B \space = \space (2.80[cos60 \hat x \space + \space sin60 \hat y]) \space \times \space (1.90[cos60 \hat x \space + \space sin60 \hat y]) \]

By **simplifying**, we get:

\[= \space -2.80 \space \times \space 1.90cos60sin60 \hat z \space – \space 2.80 \space \times \space 1.90cos60sin60 \hat z \]

\[= \space -2 \space \times \space 2.80 \space \times 1.90cos60sin60 \hat z \]

**Thus**:

\[A \space \times \space B \space = \space – 4.61 \space cm^2 \space \hat z \]

Now the **magnitude** is:

\[=\space 4.61 \space cm^2 \space \hat z \]

b) Now we have to **calculate** the **direction** for the **vector product**.

The vector product is **pointed** in the **negative direction** of the** z-axis**.

c) Now, **we have** to find the **scalar product.**

\[(\overrightarrow A \space . \space \overrightarrow B \space = \space AB \space cos \theta) \]

By **putting values**, we get:

\[= \space 20 \space cos 60 \]

\[= \space – \space 19.04 \]

d) We have to find the **scalar product**.

\[(\overrightarrow A \space . \space \overrightarrow B \space = \space AB \space cos \theta) \]

By **putting values**, we get:

\[= \space 25 \space cos 60 \]

\[= \space – \space 23.81 \]

## Numerical Answer

The **magnitude** of the **cross product** is $ 4.61 \space cm^2 \space \hat z$.

The **direction** is along the **z-axis**.

The** scalar product** is $ – \space 19.04 $.

The **scalar product** is $ – \space 23.81 $.

## Example

**Calculate** the **scalar produc**t when the **angle** is $ 30 { \circ} $, $ 90 { \circ} $ and the** vector magnitude** is $ 5 and 5 $.

First, we have to **calculate** the **scalar product** for the angle of $ 30 $ degrees.

We **know** that:

\[(\overrightarrow A \space . \space \overrightarrow B \space = \space AB \space cos \theta) \]

By **putting values**, we get:

\[= \space 25 \space cos 30 \]

\[= \space 3.85 \]

Now we have to **calculate** the **scalar product** for the angle of 90 degrees.

We **know** that:

\[(\overrightarrow A \space . \space \overrightarrow B \space = \space AB \space cos \theta) \]

By** putting values**, we get:

\[= \space 25 \space cos 90 \]

\[= \space 25 \space \times \space 0 \]

\[= \space 0 \]

Thus the** scalar product **between two vectors is equal to $ 0 $ when the angle is $ 90 $ degrees.