Vectors
1.0 Introduction
2.0 Types of Vectors
3.0 Addition of Vectors
4.0 Components of a Vector
5.0 Vector Joining Two Points
6.0 Projection of a Vector on a Line
7.0 Section Formula
8.0 Products of a Vector
9.0 Lami's Theorem
10.0 Linear Combination of Vectors
11.0 Linearly Dependent and Independent Vectors
12.0 Scalar Triple Product
13.0 Vector Triple Product
8.1 Scalar (or dot) products of two vectors
The scalar products of two non zero vectors namely $\overrightarrow a ,\overrightarrow b $, is denoted by $\overrightarrow a .\overrightarrow b $ which is defined as $$\overrightarrow a .\overrightarrow b = \left| {\overrightarrow a \,} \right|\left| {\overrightarrow b } \right|\cos \theta $$
where $\theta $ is the angle between vector $\overrightarrow a ,\overrightarrow b $, varies as $0 \leqslant \theta \leqslant \pi $. If either $\overrightarrow a or \overrightarrow b $ becomes $0$ then $\theta $ is undefined and the value of $\overrightarrow a .\overrightarrow b $ becomes $0$.
Properties
Property 1: $\overrightarrow a .\overrightarrow b $ is a real number.
Property 2: If the value of $\overrightarrow a .\overrightarrow b $ is zero for the non zero vector then it implies that the vectors are perpendicular to each other.
Note: Condition for perpendicularity of vectors $\overrightarrow a .\overrightarrow b = 0$.
Property 3: Scalar product is commutative in nature i.e., $\overrightarrow a .\overrightarrow b = \overrightarrow b .\overrightarrow a $.
Property 4: The angle between two vectors is given as $$\cos \theta = \frac{{\overrightarrow a .\overrightarrow b }}{{\left| {\overrightarrow a \,} \right|\left| {\overrightarrow b } \right|}}$$
Property 5: $\overrightarrow a .\overrightarrow a = {\left| {\overrightarrow a \,} \right|^2}$
Property 6: When vectors are parallel i.e., $\theta=0 $, then $$\begin{equation} \begin{aligned} \overrightarrow a .\overrightarrow b = \left| {\overrightarrow a \,} \right|\left| {\overrightarrow b } \right|\cos 0 \\ \overrightarrow a .\overrightarrow b = \left| {\overrightarrow a \,} \right|\left| {\overrightarrow b } \right| \\\end{aligned} \end{equation} $$
Property 7: When vectors are antiparallel i.e., $\theta = \pi $ then $$\overrightarrow a .\overrightarrow b = - \left| {\overrightarrow a \,} \right|\left| {\overrightarrow b } \right|$$
Property 8: For mutually perpendicular vectors $\widehat i,\widehat j,\widehat k$, $$\begin{equation} \begin{aligned} \widehat {i.}\widehat i = 1,\widehat j.\widehat j = 1,\widehat k.\widehat k = 1 \\ \widehat i.\widehat j = 0,\widehat j.\widehat k = 0,\widehat k.\widehat i = 0 \\\end{aligned} \end{equation} $$
Property 9: Scalar product is distributive
Proof: Let $\overrightarrow a ,\overrightarrow {b,} \overrightarrow c $ be three vectors then $$\overrightarrow a .(\overrightarrow b + \overrightarrow c ) = \overrightarrow a .\overrightarrow b + \overrightarrow a .\overrightarrow c $$
If three vectors are given in the component form $$\overrightarrow a = {a_1}\widehat i + {a_2}\widehat j + {a_3}\widehat k$$$$\overrightarrow b = {b_1}\widehat i + {b_2}\widehat j + {b_3}\widehat k$$$$\overrightarrow c = {c_1}\widehat i + {c_2}\widehat j + {c_3}\widehat k\,$$
so considering LHS, we have $$\overrightarrow a .(\overrightarrow b + \overrightarrow c )$$$$\begin{equation} \begin{aligned} ({a_1}\widehat i + {a_2}\widehat j + {a_3}\widehat k).\{ ({b_1}\widehat i + {b_2}\widehat j + {b_3}\widehat k) + ({c_1}\widehat i + {c_2}\widehat j + {c_3}\widehat k)\} \\ ({a_1}\widehat i + {a_2}\widehat j + {a_3}\widehat k).\{ ({b_1} + {c_1})\widehat i + ({b_2} + {c_2})\widehat j + ({b_3} + {c_3})\widehat k\} \\ {a_1}({b_1} + {c_1}) + {a_2}({b_2} + {c_2}) + {a_3}({b_3} + {c_3}) \\ ({a_1}{b_1} + {a_1}{c_1}) + ({a_2}{b_2} + {a_2}{c_2}) + ({a_3}{b_3} + {a_3}{c_3}) \\\end{aligned} \end{equation} $$
Now considering RHS, we have $$\overrightarrow a .\overrightarrow b + \overrightarrow a .\overrightarrow c $$$$\begin{equation} \begin{aligned} \{ ({a_1}\widehat i + {a_2}\widehat j + {a_3}\widehat k).({b_1}\widehat i + {b_2}\widehat j + {b_3}\widehat k)\} + \{ ({a_1}\widehat i + {a_2}\widehat j + {a_3}\widehat k).({c_1}\widehat i + {c_2}\widehat j + {c_3}\widehat k)\} \\ ({a_1}{b_1} + {a_2}{b_2} + {a_3}{b_3}) + ({a_1}{c_1} + {a_2}{c_3} + {a_3}{c_3}) \\ ({a_1}{b_1} + {a_1}{c_1}) + ({a_2}{b_2} + {a_2}{c_3}) + ({a_3}{b_3} + {a_3}{c_3}) \\ \\\end{aligned} \end{equation} $$
Property 10: Let $\overrightarrow a ,\overrightarrow b $ are two vectors and $\lambda $ be any scalar then
$$\lambda (\overrightarrow a .\overrightarrow b ) = (\lambda \overrightarrow a ).\overrightarrow b = \overrightarrow a .(\lambda \overrightarrow b )$$
Proof: $$\begin{equation} \begin{aligned} \lambda (\overrightarrow a .\overrightarrow b ) = \lambda ({a_1}{b_1} + {a_2}{b_2} + {a_3}{b_3}) \\ (\lambda \overrightarrow a ).\overrightarrow b = (\lambda {a_1}\widehat i + \lambda {a_2}\widehat j + \lambda {a_3}\widehat k).({b_1}\widehat i + {b_2}\widehat j + {b_3}\widehat k) \\ (\lambda \overrightarrow a ).\overrightarrow b = \lambda {a_1}{b_1} + \lambda {a_2}{b_2} + \lambda {a_3}{b_3} \\ (\lambda \overrightarrow a ).\overrightarrow b = \lambda ({a_1}{b_1} + {a_2}{b_2} + {a_3}{b_3}) \\ \lambda \overrightarrow a .(\overrightarrow b ) = \lambda ({a_1}\widehat i + {a_2}\widehat j + {a_3}\widehat k).({b_1}\widehat i + {b_2}\widehat j + {b_3}\widehat k) \\ \overrightarrow a .(\lambda \overrightarrow b ) = ({a_1}\widehat i + {a_2}\widehat j + {a_3}\widehat k).(\lambda {b_1}\widehat i + \lambda {b_2}\widehat j + \lambda {b_3}\widehat k) \\ \overrightarrow a .(\lambda \overrightarrow b ) = {a_1}\lambda {b_1} + {a_2}\lambda {b_2} + {a_3}\lambda {b_3} \\ \overrightarrow a .(\lambda \overrightarrow b ) = \lambda {a_1}{b_1} + \lambda {a_2}{b_2} + \lambda {a_3}b \\\end{aligned} \end{equation} $$
