# 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 $$\vec a$$

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

Thus,

$\hat i \cdot \hat i = \hat j \cdot \hat j = \hat k \cdot \hat k = 1$

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

\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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