Properties Of Vector Dot Product

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From the definition of the dot product, we can make certain useful observations about its properties.

(i) The angle \(\theta\; between \;two\; vectors\; \vec a\,\,{\text{and}}\,\,\vec b\) is given by

\[\cos \theta  = \frac{{\vec a \cdot \vec b}}{{\left| {\vec a} \right|\;\left| {\vec b} \right|}}\]

(ii) \(\left| {\vec a \cdot \vec b} \right| \leqslant \left| {\vec a} \right|\;\left| {\vec b} \right|\), the equality holding only if  \(\theta  = 0\;\;{\text{or}}\;\;\pi \)

(iii) The projection of  \(\vec a\) on \(\vec b\) is

\[{p_{ab}} = \frac{{\vec a \cdot \vec b}}{{\left| {\vec b} \right|}} = \vec a \cdot \left( {\frac{{\vec b}}{{\left| {\vec b} \right|}}} \right) = \vec a \cdot \widehat b\]

(iv) The projection of \(\vec b\;\;{\text{on}}\;\;\vec a\)  is

\[{p_{ba}} = \frac{{\vec a \cdot \vec b}}{{\left| {\vec a} \right|}} = \left( {\frac{{\vec a}}{{\left| {\vec a} \right|}}} \right) \cdot \vec b = \widehat a \cdot \vec b\]

(v) Scalar product is commutative, i.e,

\[\vec a \cdot \vec b = \vec b \cdot \vec a\]

(vi) Scalar product is distributive, i.e

\[\begin{align}&\qquad\;\;\vec a \cdot \left( {\vec b + \vec c} \right) = \vec a \cdot \vec b + \vec a \cdot \vec c \hfill \\& and\,\,\,\,\,\left( {\vec a + \vec b} \right) \cdot \vec c = \vec a \cdot \vec c + \vec b \cdot \vec c \hfill \\ \end{align} \]

(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 \(\vec a\)

\[\boxed{\vec a \cdot \vec a = {{\left| {\vec a} \right|}^2}}\]


\[\hat i \cdot \hat i = \hat j \cdot \hat j = \hat k \cdot \hat k = 1\]

(viii)             \[\left |{\vec a \pm \vec b} \right| =(\vec a \pm \vec b) \cdot(\vec a \pm \vec b)\]

\[\begin{align} = {\left| {\vec a} \right|^2} + {\left| {\vec b} \right|^2} \pm 2(\vec a \cdot \vec b) \hfill \\ (\vec a + \vec b)(\vec a - \vec b) = {\left| {\vec a} \right|^2} - {\left| {\vec b} \right|^2} \hfill \\ \end{align} \]

(ix) This property is very important. If two vectors  \(\vec a\,\,{\text{and}}\,\,\vec b\) have been specified in rectangular form, i.e.,

\[\vec a = {a_1}\,\hat i + {a_2}\,\hat j + {a_3}\,\hat k\;\;{\text{and}}\;\;\vec b = {b_1}\,\hat i + {b_2}\,\hat j + {b_3}\,\hat k\,\,\,then\]

we have,

\[\begin{align}&\qquad\quad\vec a \cdot \vec b = \left( {{a_1}\hat i + {a_2}\hat j + {a_3}\hat k} \right)\left( {{b_1}\hat i + {b_2}\hat j + {b_3}\hat k} \right) \hfill \\\\&\qquad\qquad\;\;= {a_1}\,{b_1} + {a_2}\,{b_2} + {a_3}\,{b_3}\;\;\;\;\;\;\;\quad\left \{ {Using{\text{ }}properties{\text{ }}\left( {vi} \right){\text{ }}and{\text{ }}\left( {vii} \right)} \right\} \hfill \\\\&\Rightarrow\; \,\,\,\,\boxed{\vec a \cdot \vec b = {a_1}\,{b_1} + {a_2}\,{b_2} + {a_3}\,{b_3}} \hfill \\ \end{align} \]

The angle \(\theta\) between the two vectors will be given by \(\begin{align}cos \theta = \frac{{\vec a \cdot \vec b}}{{\left| {\vec a} \right|\;\left| {\vec b} \right|}}:\end{align}\)

\(\begin{align} \Rightarrow  \qquad \cos \theta  = \frac{{{a_1}{\kern 1pt} \,{b_1} + {a_2}\,{b_2} + {a_3}\,{b_3}}}{{\sqrt {a_1^2 + a_2^2 + a_3^2} \sqrt {b_1^2 + b_2^2 + b_3^2} }}\end{align}\)

(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 \(\vec a\,\,{\text{and}}\,\,\vec b\) . If \(\vec r \cdot \vec a = 0\) and  \(\vec r \cdot \vec b = 0,\) this would imply that  \(\vec r\) is perpendicular to both \(\vec a\,\,{\text{and}}\,\,\vec b\) . This can only happen if \(\vec a\,\,{\text{and}}\,\,\vec b\) 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 \(\vec r\) that is perpendicular to each of \(\vec a,\,\,\vec b\,\,{\text{and}}\,\vec c\)  which can only happen if \(\vec a,\,\,\vec b\,\,{\text{and}}\,\vec c\) 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

\[\boxed {\begin{align}\vec r &= (\vec r \cdot \hat a)\hat a + (\vec r \cdot \hat b)\hat b + (\vec r \cdot \hat c)\hat c  \\& = \left( {\frac{{\vec r \cdot \vec a}}{{{{\left| {\vec a} \right|}^2}}}} \right)\vec a + \left( {\frac{{\vec r \cdot \vec b}}{{{{\left| {\vec b} \right|}^2}}}} \right)\vec b + \left( {\frac{{\vec r \cdot \vec c}}{{{{\left| {\vec c} \right|}^2}}}} \right)\vec c \\ \end{align}}\]

This representation is of significant importance and you must understand how it comes about.

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