# For the two vectors in the figure (Figure 1) , find the magnitude of the vector product

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

– 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:

$A \space \times \space B \space = \space ||A || \space || B || \space sin \theta n$

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:

$= \space 25 \space cos 60$

$= \space – \space 23.81$

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 product 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.