## Friday, 8 August 2014

### CHAPTER 17-Properties of Dot Product

From the definition of the dot product, we can make certain useful observations about its properties.
(i) The angle $\theta$ between two vectors $\vec a$ and $\vec b$ is given by
 $\cos \theta = \dfrac{{\vec a \cdot \vec b}}{{\left| {\vec a} \right|\;\left| {\vec b} \right|}}$
(ii) $\left| {\vec a \cdot \vec b} \right| \le \left| {\vec a} \right|\;\left| {\vec b} \right|$, the equality holding only if $\theta = 0$ or $\pi$
(iii) The projection of $\vec a$ on $\vec b$ is
 ${p_{ab}} = \dfrac{{\vec a \cdot \vec b}}{{\left| {\vec b} \right|}} = \vec a \cdot \left( {\dfrac{{\vec b}}{{\left| {\vec b} \right|}}} \right) = \vec a \cdot \widehat b$
(iv) The projection of $\vec b$ on $\vec a$ is
 ${p_{ba}} = \dfrac{{\vec a \cdot \vec b}}{{\left| {\vec a} \right|}} = \left( {\dfrac{{\vec a}}{{\left| {\vec a} \right|}}} \right) \cdot \vec b = \widehat a \cdot \vec b$
(v) Scalar product is commutative i.e.,
 $\vec a \cdot \vec b = \vec b \cdot \vec a$
(vi) Scalar product is distributive i.e.,
 $\vec a \cdot \left( {\vec b + \vec c} \right) = \vec a \cdot \vec b + \vec a \cdot \vec c$ and $\left( {\vec a + \vec b} \right) \cdot \vec c = \vec a \cdot \vec c + \vec b \cdot \vec c$
(vi) The scalar product of two vectors is zero if and only if the two vectors are perpendicular.
This also gives
 $\hat i \cdot \hat j = \hat j \cdot \hat i = \hat i \cdot \hat k = \hat k \cdot \hat i = \hat j \cdot \hat k = \hat k \cdot \hat j = 0$
(vii) For any vector $\vec a$
${\vec a \cdot \vec a = {{\left| {\vec a} \right|}^2}}$
Thus,
 $\hat i \cdot \hat i = \hat j \cdot \hat j = \hat k \cdot \hat k = 1$
(viii) ${\left| {\vec a \pm \vec b} \right|^2} = (\vec a \pm \vec b) \cdot (\vec a \pm \vec b)$
 $= {\left| {\vec a} \right|^2} + {\left| {\vec b} \right|^2} \pm 2(\vec a \cdot \vec b)$ $(\vec a + \vec b)(\vec a - \vec b) = {\left| {\vec a} \right|^2} - {\left| {\vec b} \right|^2}$
(ix) This property is very important. If two vectors $\vec a$ and $\vec b$ have been specified in rectangular form
 $\vec a = {a_1}\,\hat i + {a_2}\,\hat j + {a_3}\,\hat k$ and $\vec b = {b_1}\,\hat i + {b_2}\,\hat j + {b_3}\,\hat k$ then
We have
 $\vec a \cdot \vec b = \left( {{a_1}\hat i + {a_2}\hat j + {a_3}\hat k} \right)\left( {{b_1}\hat i + {b_2}\hat j + {b_3}\hat k} \right)$ $= {a_1}\,{b_1} + {a_2}\,{b_2} + {a_3}\,{b_3}$ Using properties $(vi)$ and $(vii)$ $\Rightarrow\,\,\,\,{\vec a \cdot \vec b = {a_1}\,{b_1} + {a_2}\,{b_2} + {a_3}\,{b_3}}$
The angle $\theta$ between the two vectors will be given by $\cos \theta = \dfrac{{\vec a \cdot \vec b}}{{\left| {\vec a} \right|\;\left| {\vec b} \right|}}:$
 $\Rightarrow \,\,\,\, \cos \theta = \dfrac{{{a_1}{\kern 1pt} \,{b_1} + {a_2}\,{b_2} + {a_3}\,{b_3}}}{{\sqrt {a_1^2 + a_2^2 + a_3^2} \sqrt {b_1^2 + b_2^2 + b_3^2} }}$
(x) The direction cosines $l$$m$$n$ of a vector $\vec a$ will be given by
 $l = \hat a \cdot \hat i,\;\;\;m = \hat a \cdot \hat j,\;\;\;\;n = \hat a \cdot \hat k$
(xi) Let $\vec r$ be a vector coplanar with the vectors $\vec a$ and $\vec b$. If $\vec r \cdot \vec a = 0$ and $\vec r \cdot \vec b = 0,$ this would imply that $\vec r$ is perpendicular to both $\vec a$ and $\vec b$. This can only happen if $\vec a$ and $\vec b$ are collinear.
Analogously, let $\vec r$ be an arbitrary vector and $\vec a,\;\;\vec b,\;\;\vec c$ be three vectors such that
 $\vec r \cdot \vec a = \vec r \cdot \vec b = \vec r \cdot \vec c = 0$
This means that $\vec r$ is perpendicular to each of $\vec a$$\vec b$ and $\vec c$ which can only happen if $\vec a,\;\vec b$ and $\vec c$ are coplanar.
(xii) Let $\vec a,\;\;\vec b,\;\;\vec c$ be three non-coplanar vectors. We’ve already discussed that $\vec a,\;\;\vec b,\;\;\vec c$ can form a basis for $3 - D$ space. Any vector $\vec r$ can be written in this basis as
$\begin{array}{l} \vec r = (\vec r \cdot \hat a)\hat a + (\vec r \cdot \hat b)\hat b + (\vec r \cdot \hat c)\hat c\\ \,\,\,\, = \left( {\dfrac{{\vec r \cdot \vec a}}{{{{\left| {\vec a} \right|}^2}}}} \right)\vec a + \left( {\dfrac{{\vec r \cdot \vec b}}{{{{\left| {\vec b} \right|}^2}}}} \right)\vec b + \left( {\dfrac{{\vec r \cdot \vec c}}{{{{\left| {\vec c} \right|}^2}}}} \right)\vec c \end{array}$
This representation is of significant importance and you must understand how it comes about.