# Product of Two Vectors – Scalar Product & Vector Product

Vector Algebra / Tuesday, October 29th, 2019

# Product of Two Vectors

The product of two vectors is defined in two ways, scalar product and vector product. A vector being a physical quantity having magnitude as well as direction, the process by which product of two or more vectors is formed, will obviously be different from usual operation of multiplication in arithmetic. Both scalar and vector product of two vectors obey certain law, which will be discussed here.

## Scalar Product or Dot Product

Scalar product of two vectors α and β, denoted by α.β (read α dot β), is a scalar quantity defined as the product of the magnitudes of α and β and the cosine of the angle between them.

If θ (0 ≤ θ ≤ π) is angle between α and β, then

$\overrightarrow{\alpha }.\overrightarrow{\beta }=\left| \overrightarrow{\alpha } \right|\left| \overrightarrow{\beta } \right|\cos \theta$

 Example

Let α and β be two vectors such that| α | = 5 and | β | = 3 and angle between them is 30o. Then,

$\overrightarrow{\alpha }.\overrightarrow{\beta }=\left| \overrightarrow{\alpha } \right|\left| \overrightarrow{\beta } \right|\cos {{30}^{o}}=5\times 3\times \frac{\sqrt{3}}{2}=\frac{15\sqrt{3}}{2}$

Note: The scalar product between two vectors is a scalar, not a vector.

## Properties of Scalar Product

The following laws are valid for any vectors α, β, and γ:

$1)\overrightarrow{\alpha }.\overrightarrow{\beta }=\overrightarrow{\beta }.\overrightarrow{\alpha }\left( \text{Commutative law for scalar product} \right)$

$2)\overrightarrow{\alpha }.\left( \overrightarrow{\beta }+\overrightarrow{\gamma } \right)=\overrightarrow{\alpha }.\overrightarrow{\beta }+\overrightarrow{\alpha }.\overrightarrow{\gamma }\left( \text{Distributive law} \right)$

$3)x\left( \overrightarrow{\alpha }.\overrightarrow{\beta } \right)=\left( x\overrightarrow{\alpha } \right).\overrightarrow{\beta }=\overrightarrow{\alpha }.\left( x\overrightarrow{\beta } \right)=\left( \overrightarrow{\alpha }.\overrightarrow{\beta } \right)x\left( \text{where x is a scalar} \right)$

$4)\overrightarrow{\alpha }.\overrightarrow{\alpha }={{\left| \overrightarrow{\alpha } \right|}^{2}}$

$5)\overrightarrow{\alpha }.\overrightarrow{\beta }=0~\text{if and only if}\overrightarrow{\alpha }~\text{and}~~\overrightarrow{\beta }~~are~~perpendicular$

$to~~each~~other(\overrightarrow{\alpha },\overrightarrow{\beta }\ne 0)$

$6)If~~\overrightarrow{\alpha }={{x}_{1}}\hat{i}+{{y}_{1}}\hat{j}+{{z}_{1}}\hat{k}~~and~~\overrightarrow{\beta }={{x}_{2}}\hat{i}+{{y}_{2}}\hat{j}+{{z}_{2}}\hat{k}~~then$

$\overrightarrow{\alpha }.\overrightarrow{\beta }={{x}_{1}}{{x}_{2}}+{{y}_{1}}{{y}_{2}}+{{z}_{1}}{{z}_{2}}.$

$7)\overrightarrow{\alpha }.\overrightarrow{\beta }=\left| \overrightarrow{\alpha } \right|~~projection~~of~~\overrightarrow{\beta }~~on~~the~~axis~~of~~\overrightarrow{\alpha }$

$\left( i.e.,~~the~~line~~axis~~on~~which~~\overrightarrow{\alpha ~}~~lies \right)$

## Projection of Vector

Without loss of generality we take α and β have common initial point, say A.

In the above figure g is the axis of the vector α. B is the foot of perpendicular on g drawn from the terminal point of β. Then |AB| = Projection of β on g.

$Obviously~~\cos \theta =\frac{\left| AB \right|}{\left| \overrightarrow{\beta } \right|}$

$\Rightarrow \left| AB \right|=\left| \overrightarrow{\beta } \right|\cos \theta$

$\therefore \text{Projection of}~~\overrightarrow{\beta }~~on~~g=\left| \overrightarrow{\beta } \right|\cos \theta$

$\Rightarrow \left| \overrightarrow{\alpha } \right|~~\text{Projection of}~~\overrightarrow{\beta }~~on~~g=\left| \overrightarrow{\alpha } \right|\left| \overrightarrow{\beta } \right|\cos \theta =\overrightarrow{\alpha }.\overrightarrow{\beta }$

$\therefore \overrightarrow{\alpha }.\overrightarrow{\beta }=\left| \overrightarrow{\alpha } \right|~~projection~~of~~\overrightarrow{\beta }~~on~~the~~axis~~of~~\overrightarrow{\alpha }$

 Theorem

If an axis g makes angle α, β and γ with the coordinate axes respectively (i.e., cosα, cosβ, cosγ are direction cosine of g) then projection of a vector ρ = xi + yj + xk on g is given by

$\text{Projection of}~~\overrightarrow{\rho }~~on~~g=x\cos \alpha +y\cos \beta +z\cos \gamma$

## Vector Product or Cross Product

The vector product of α and β, denoted by

$\overrightarrow{\alpha }\times \overrightarrow{\beta }$

(read α cross β) is defined as

$\overrightarrow{\alpha }\times \overrightarrow{\beta }=\left| \overrightarrow{\alpha } \right|\left| \overrightarrow{\beta } \right|\sin \theta \hat{n}$

Where θ (0 ≤ θ ≤ π) is angle between α and β and n is a unit vector perpendicular to the plane of such α and β that α, β and n form a right handed system i.e., n is along the direction in which a screw head would advance relative to a rotation from α to β.

In the above figure we show the two different directions of n depending on the position of α and β.

Note: α X β is a vector which is perpendicular to both of α and β.

 Example

If α and β be two vectors such that| α | = 5 and | β | = 3 and angle between them is 30o. Then,

$\overrightarrow{\alpha }\times \overrightarrow{\beta }=\left| \overrightarrow{\alpha } \right|\left| \overrightarrow{\beta } \right|\sin {{30}^{0}}\hat{n}=15\times \frac{1}{2}\hat{n}=\frac{15}{2}\hat{n}$

## Properties of Vector Product

The following properties hold for any vectors α, β and γ.

$1)\overrightarrow{\alpha }\times \overrightarrow{\beta }=-\overrightarrow{\beta }\times \overrightarrow{\alpha }~~\left( commutative~~law~~does~~not~~hold \right)$

$2)\overrightarrow{\alpha }\times \left( \overrightarrow{\beta }+\overrightarrow{\gamma } \right)=\overrightarrow{\alpha }\times \overrightarrow{\beta }+\overrightarrow{\alpha }\times \overrightarrow{\gamma }~~\left( Distributive~~law \right)$

$3)x\left( \overrightarrow{\alpha }\times \overrightarrow{\beta } \right)=\left( x\overrightarrow{\alpha } \right)\times \overrightarrow{\beta }=\overrightarrow{\alpha }\times \left( x\overrightarrow{\beta } \right)=\left( \overrightarrow{\alpha }\times \overrightarrow{\beta } \right)x,~~where~~x~~is~~a~~scalar$

$4)\overrightarrow{\alpha }\times \overrightarrow{\beta }=\overrightarrow{0}~~if~~and~~if~~\overrightarrow{\alpha }~~and~~\overrightarrow{\beta }~~are~~parallel$

$5)\overrightarrow{\alpha }\times \overrightarrow{\alpha }=\overrightarrow{0}~~$

$6)\hat{i}\times \hat{j}=\hat{k},\hat{j}\times \hat{k}=\hat{i},\hat{k}\times \hat{i}=\hat{j}$

$7)\left| \overrightarrow{\alpha }\times \overrightarrow{\beta } \right|=~~\text{Area of a parallelogram with adjacent sides}~~\overrightarrow{\alpha ~}~~and~~\overrightarrow{\beta }.$

 Example 01

Find the projection (or component) of the vector α on the vector β, where

$\overrightarrow{\alpha }=\hat{i}-2\hat{j}+\hat{k},\overrightarrow{\beta }=4\hat{i}-4\hat{j}+7\hat{k}$

Solution:

$\overrightarrow{\alpha }.\overrightarrow{\beta }=1\times 4+\left( -2 \right)\times \left( -4 \right)+1\times 7=4+8+7=19$

$\left| \overrightarrow{\beta } \right|=\sqrt{{{4}^{2}}+{{\left( -4 \right)}^{2}}+{{7}^{2}}}=\sqrt{81}=9$

$\text{Now},~~\overrightarrow{\alpha }.\overrightarrow{\beta }=\left| \overrightarrow{\beta } \right|~~~\text{projection of}~~\overrightarrow{\alpha }~~on~~\overrightarrow{\beta }$

$\therefore \text{projection of}~~\overrightarrow{\alpha }~~on~~\overrightarrow{\beta }=\frac{19}{9}$

 Example 02

Find the value of

$\overrightarrow{\alpha }.\overrightarrow{\beta }+\overrightarrow{\beta }.\overrightarrow{\gamma }+\overrightarrow{\gamma }.\overrightarrow{\alpha }~~where~~\left| \overrightarrow{\alpha } \right|=\left| \overrightarrow{\beta } \right|=\left| \overrightarrow{\gamma } \right|~~and~~\overrightarrow{\alpha }+\overrightarrow{\beta }+\overrightarrow{\gamma }=\overrightarrow{0}.$

Solution:

Given,

$\overrightarrow{\alpha }+\overrightarrow{\beta }+\overrightarrow{\gamma }=\overrightarrow{0}$

$\Rightarrow \left( \overrightarrow{\alpha }+\overrightarrow{\beta }+\overrightarrow{\gamma } \right).\overrightarrow{\alpha }=\overrightarrow{0}.\overrightarrow{\alpha }$

$\Rightarrow \overrightarrow{\alpha }.\overrightarrow{\alpha }+\overrightarrow{\beta }.\overrightarrow{\alpha }+\overrightarrow{\gamma }.\overrightarrow{\alpha }=0$

$\Rightarrow {{\left| \overrightarrow{\alpha } \right|}^{2}}+\overrightarrow{\beta }.\overrightarrow{\alpha }+\overrightarrow{\gamma }.\overrightarrow{\alpha }=0$

$\Rightarrow {{\left( 1 \right)}^{2}}+\overrightarrow{\beta }.\overrightarrow{\alpha }+\overrightarrow{\gamma }.\overrightarrow{\alpha }=0$

$\Rightarrow \overrightarrow{\beta }.\overrightarrow{\alpha }+\overrightarrow{\gamma }.\overrightarrow{\alpha }=-1$

Similarly we get,

$\overrightarrow{\beta }.\overrightarrow{\gamma }+\overrightarrow{\gamma }.\overrightarrow{\alpha }=-1~~and~\overrightarrow{\beta }.\overrightarrow{\gamma }+\overrightarrow{\beta }.\overrightarrow{\alpha }=-1$

$2\left( \overrightarrow{\alpha }.\overrightarrow{\beta }+\overrightarrow{\beta }.\overrightarrow{\gamma }+\overrightarrow{\gamma }.\overrightarrow{\alpha }~ \right)=-3$

$\Rightarrow \overrightarrow{\alpha }.\overrightarrow{\beta }+\overrightarrow{\beta }.\overrightarrow{\gamma }+\overrightarrow{\gamma }.\overrightarrow{\alpha }~=-\frac{3}{2}$

 Example 03

$If~~\left| \overrightarrow{\alpha }+\overrightarrow{\beta } \right|=\left| \overrightarrow{\alpha }-\overrightarrow{\beta } \right|,~~prove~~that~~\overrightarrow{\alpha }~~is~~perpendicular~~to~~\overrightarrow{\beta }.$

Solution:

$Let~~\overrightarrow{\alpha }=\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right),~~\overrightarrow{\beta }=\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)$

$\therefore \overrightarrow{\alpha }+\overrightarrow{\beta }=\left( {{x}_{1}}+{{x}_{2}},{{y}_{1}}+{{y}_{2}},{{z}_{1}}+{{z}_{2}} \right)$

$\overrightarrow{\alpha }-\overrightarrow{\beta }=\left( {{x}_{1}}-{{x}_{2}},{{y}_{1}}-{{y}_{2}},{{z}_{1}}-{{z}_{2}} \right)$

$\therefore ~\left| \overrightarrow{\alpha }+\overrightarrow{\beta } \right|=\sqrt{{{\left( {{x}_{1}}+{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}+{{y}_{2}} \right)}^{2}}+{{\left( {{z}_{1}}+{{z}_{2}} \right)}^{2}}}$

$and~~\left| \overrightarrow{\alpha }-\overrightarrow{\beta } \right|=\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}+{{\left( {{z}_{1}}-{{z}_{2}} \right)}^{2}}}$

$\because \left| \overrightarrow{\alpha }+\overrightarrow{\beta } \right|=\left| \overrightarrow{\alpha }-\overrightarrow{\beta } \right|$

$\Rightarrow \sqrt{{{\left( {{x}_{1}}+{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}+{{y}_{2}} \right)}^{2}}+{{\left( {{z}_{1}}+{{z}_{2}} \right)}^{2}}}=\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}+{{\left( {{z}_{1}}-{{z}_{2}} \right)}^{2}}}$

$\Rightarrow 2{{x}_{1}}{{x}_{2}}+2{{y}_{1}}{{y}_{2}}+2{{z}_{1}}{{z}_{2}}=-2{{x}_{1}}{{x}_{2}}-2{{y}_{1}}{{y}_{2}}-2{{z}_{1}}{{z}_{2}}$

$\Rightarrow 4\left( {{x}_{1}}{{x}_{2}}+{{y}_{1}}{{y}_{2}}+{{z}_{1}}{{z}_{2}} \right)=0$

$\Rightarrow {{x}_{1}}{{x}_{2}}+{{y}_{1}}{{y}_{2}}+{{z}_{1}}{{z}_{2}}=0$

$\Rightarrow \overrightarrow{\alpha }.\overrightarrow{\beta }=0$

$\Rightarrow \overrightarrow{\alpha }~~is~~perpendicular~~to~~\overrightarrow{\beta }$