The Scalar and Vector Product

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Scalar and Vector product

There are two different ways in which vectors can be multiplied: the scalar and the vector product.

Scalar Product

Let’s suppose that we have two vectors $\displaystyle \overrightarrow{a}$and $\displaystyle \overrightarrow{b}$ shown below with the same direction and also the angle between them labeled ϴ.

Definition: The scalar product of the vector $\displaystyle \overrightarrow{a}$and $\displaystyle \overrightarrow{b}$ is the product of the magnitudes of these vectors and the cosine of the angle between them.

$\displaystyle \vec{a}\cdot \vec{b}=\left| \overrightarrow{a} \right|\cdot \left| \overrightarrow{b} \right|\cos \theta$

The scalar product of those two vectors is denoted by a thick dot and is always a scalar quantity.

Properties of scalar product:

1. $\displaystyle \overrightarrow{a}\cdot \vec{b}=\vec{b}\cdot \overrightarrow{a}$

2. If $ \displaystyle \overrightarrow{a}$ and $\displaystyle \overrightarrow{b}$ are two non-zero vectors, then $\displaystyle \overrightarrow{a}\cdot \vec{b}=0\Leftrightarrow \overrightarrow{a}\bot \vec{b}$

3. If $\displaystyle \overrightarrow{a}\cdot \vec{b}=\vec{b}\cdot \overrightarrow{a}=ab$ and $\displaystyle \vec{b}$ are two vectors parallel to each other.

4. $\displaystyle \overrightarrow{a}(\vec{b}+\overrightarrow{c})=\overrightarrow{a}\cdot \vec{b}+\overrightarrow{a}\cdot \overrightarrow{c}$

5. If ϴ=0 then $\displaystyle \overrightarrow{a}\cdot \vec{b}=\left| {\overrightarrow{a}} \right|\left| {\vec{b}} \right|$ and if $\displaystyle \overrightarrow{a}=\vec{b}$ then $\displaystyle \overrightarrow{a}\cdot \overrightarrow{a}={{\left| \overrightarrow{a} \right|}^{2}}$

 

6. If $\displaystyle \overrightarrow{a}$ and $\displaystyle \overrightarrow{b}$ are any to vectors and $\displaystyle \lambda $ is a scalar then $\displaystyle (\lambda \overrightarrow{a})\cdot \vec{b}=\lambda (\overrightarrow{a}\overrightarrow{{\cdot b}})=\overrightarrow{a}(\lambda \vec{b})$

Example 1: Find the product $\displaystyle \overrightarrow{a}\cdot \overrightarrow{b}$ where $\displaystyle \left| \overrightarrow{a} \right|=\sqrt{3}$ and $\displaystyle \left| \overrightarrow{b} \right|=2$ and the angle between $\displaystyle \overrightarrow{a}$ and $\displaystyle \overrightarrow{b}$ is 45°.

Solution: From the definition we know that the scalar product of two vectors is:

$\displaystyle \overrightarrow{a}\cdot \vec{b}=\left| {\overrightarrow{a}} \right|\cdot \left| {\vec{b}} \right|\cos \theta $

In our example we have:

$\displaystyle \left| {\overrightarrow{a}} \right|=\sqrt{3}$

$\displaystyle \left| {\overrightarrow{b}} \right|=2$

$\displaystyle \theta $ = 45°

Replacing the values we have in our formula, we get:

$\displaystyle \overrightarrow{a}\cdot \vec{b}=\left| \overrightarrow{a} \right|\cdot \left| {\vec{b}} \right|\cos \theta $

$\displaystyle \overrightarrow{a}\cdot \vec{b}=\sqrt{3}\cdot 2\cos {{45}^{{^{{}^\circ }}}}$

$\displaystyle \overrightarrow{a}\cdot \vec{b}=2\sqrt{3}\cdot \frac{{\sqrt{2}}}{2}=\sqrt{6}$

So, the scalar product of these two vectors is the scalar value $\displaystyle \sqrt{6}$.

Using the scalar product to find the angle between two vectors

The angle between two vectors is found from the formula: $\displaystyle \cos \alpha =\frac{{\overrightarrow{a}\cdot \vec{b}}}{{\left| a \right|\cdot \left| b \right|}}$

Example 2: Find the angle between two vectors $\displaystyle\overrightarrow{a}$ and $\displaystyle {\vec{b}}$, where the magnitudes are $\displaystyle \left| \overrightarrow{a} \right|=2$  and $\displaystyle \left| {\vec{b}} \right|=4$ with product scalar respectively $\displaystyle \overrightarrow{a}\cdot \vec{b}=4$

Solution: From the definition we know that the scalar product of two vectors is:

$\displaystyle \overrightarrow{a}\cdot \vec{b}=\left| \overrightarrow{a} \right|\cdot \left| {\vec{b}} \right|\cos \theta $

In our example we have:

$\displaystyle \left| \overrightarrow{a} \right|=2$

$\displaystyle \left| {\vec{b}} \right|=4$

$\displaystyle \overrightarrow{a}\cdot \vec{b}=4$

 

Replacing the values we have in our formula, we get:

$\displaystyle \cos \alpha =\frac{\vec{a}\cdot \vec{b}}{\left| a \right|\cdot \left| b \right|}$

$\displaystyle \cos \alpha =\frac{4}{2\cdot 4}=\frac{1}{2}$

$\displaystyle \alpha ={{60}^{\circ }}$

So, the angle between those two vectors is 60°.

Vector product

Let`s suppose that we have two vectors $\displaystyle \overrightarrow{a}$and $\displaystyle \overrightarrow{b}$ shown below with the same direction and also the angle between them labeled ϴ. We label the unit vector $\displaystyle \overrightarrow{n}$ the is perpendicular with the plane of the vectors $\displaystyle \overrightarrow{a}$ and $\displaystyle \overrightarrow{b}$

DefinitionThe cross product of the vector $\displaystyle \overrightarrow{a}$and $\displaystyle \overrightarrow{b}$ is the product of the magnitudes of these vectors and the sinuses of the angle between them.

 
$\displaystyle \overrightarrow{a}\times \vec{b}=\left| {\overrightarrow{a}} \right|\cdot \left| {\overrightarrow{b}} \right|\sin \theta \vec{n}$

*The cross product of those two vectors is denoted by a cross symbol and is always a vector

The magnitude of the vector axb is:
$\displaystyle \left| {\overrightarrow{a}} \right.\times \left. {\vec{b}} \right|=\left| {\overrightarrow{a}} \right|\cdot \left| {\vec{b}} \right|\sin \theta \vec{n}$

Properties of cross product:

1. $ \displaystyle \left( {\overrightarrow{a}\times \vec{b}} \right)=-\left( {\vec{b}\times \overrightarrow{a}} \right)$

2. $ \displaystyle k\overrightarrow{a}\times m\vec{b}=km\left( {\overrightarrow{a}} \right.\times \left. {\vec{b}} \right)$

3. $ \displaystyle \left( {\overrightarrow{a}+} \right.\left. {\vec{b}} \right)\times \overrightarrow{c}=\left( {\overrightarrow{a}} \right.\times \left. {\overrightarrow{c}} \right)+\left( {\vec{b}} \right.\times \left. {\overrightarrow{c}} \right)$

4. $\displaystyle \overrightarrow{a}\times \vec{b}=0\Leftrightarrow \overrightarrow{a}\parallel \vec{b}$

Example 3: Find the magnitude of the vector axb if $\displaystyle \left| {\overrightarrow{a}} \right|=2$ and $\displaystyle \left| {\overrightarrow{b}} \right|=4$ and the angle between the vectors is $\displaystyle \theta ={{45}^{\circ }}$.

Solution: We know that the magnitude of the vector axb is:

$\displaystyle \left| {\overrightarrow{a}} \right.\times \left. {\vec{b}} \right|=\left| {\overrightarrow{a}} \right|\cdot \left| {\vec{b}} \right|\sin \theta \overrightarrow{n}$

In our example we have:

$\displaystyle \left| {\overrightarrow{a}} \right|=2$

$\displaystyle \left| {\overrightarrow{b}} \right|=4$

$\displaystyle \theta ={{45}^{\circ }}$

Replacing the values we have in our formula, we get:

$\displaystyle \left| {\overrightarrow{a}} \right.\times \left. {\vec{b}} \right|=\left| {\overrightarrow{a}} \right|\cdot \left| {\vec{b}} \right|\sin \theta $

$ \displaystyle \left| {\overrightarrow{a}} \right.\times \left. {\vec{b}} \right|=2\cdot 4\sin {{45}^{{}^\circ }}$

$\displaystyle \left| {\overrightarrow{a}} \right.\times \left. {\vec{b}} \right|=2\cdot 4\cdot \frac{{\sqrt{2}}}{2}$

$\displaystyle \left| {\overrightarrow{a}} \right.\times \left. {\vec{b}} \right|=4\sqrt{2}$

So, the magnitude of the vector a x b is $\displaystyle \overrightarrow{a}\times \vec{b}=\overrightarrow{{27,65}}$

Example 4: Find the vector product of two vectors if the scalar product is 21 and the magnitudes of these vectors are 5 and 7.

Solution: From the definition we know that the scalar product of two vectors a and b is:

$\displaystyle \overrightarrow{a}\cdot \vec{b}=\left| \overrightarrow{a} \right|\cdot \left| {\vec{b}} \right|\cos \theta $

And the vector product is:

$ \displaystyle \vec{a}\times \vec{b}=a\cdot b\sin \theta \vec{n}$

In our example we have:

$\displaystyle \left| \overrightarrow{a} \right|=5$

$\displaystyle \left| \overrightarrow{b} \right|=7$

$ \displaystyle \overrightarrow{a}\cdot \vec{b}=21$

To find the vector product we need to find the angle between the two vectors.

$\displaystyle \cos \alpha =\frac{\overrightarrow{a}\cdot \overrightarrow{b}}{\left| a \right|\cdot \left| b \right|}$

$\displaystyle \cos \alpha =\frac{21}{5\cdot 7}=\frac{3}{5}=0,6$

$ \displaystyle \alpha ={{\cos }^{{-1}}}(0,6)$

$ \displaystyle \alpha ={{53}^{\circ }}$

Replacing the values we have in our formula, we get:

$\displaystyle \overrightarrow{a}\times \vec{b}=\left| {\overrightarrow{a}} \right|\cdot \left| {\overrightarrow{b}} \right|\sin \theta \vec{n}$

$ \displaystyle \overrightarrow{a}\times \vec{b}=5\cdot 7\sin {{53}^{{}^\circ }}$

So, the vector product of these two vectors is the vector  $ \displaystyle \overrightarrow{{27,65}}$

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