Scalar (or Dot) Product of Two Vectors

SCALAR (OR DOT) PRODUCT OF TWO VECTORS

The scalar product of $\bar{a}$and $\bar{b}$, written as $\bar{a}\cdot \bar{b}$, is defined to be the $|\bar{a}||\bar{b}|$cos$\theta$ where $\theta$ is the angle between the vectors $\bar{a}$and $\bar{b}$ i.e. $\bar{a}\cdot \bar{b}$ = abcos$\theta$.

Geometrical Interpretation:

$\bar{a}.\cdot \bar{b}$ is the product of length of one vector and length of the projection of the other vector in the direction of former.

$\bar{a}\cdot \bar{b}=\left|\bar{a}\right|$. Projection of $\left|\bar{b}\right|$ in direction of $\bar{a}$ =$\left|\bar{b}\right|$.Projection of $\left|\bar{a}\right|$ in direction of $\bar{b}$

Properties:

$\bullet$ $\bar{a}\cdot \bar{a}=|\bar{a}{|}^2=a^2$ $\Rightarrow$ $\hat{i}\cdot \hat{i}=\hat{j}\cdot \hat{j}=\hat{k}\cdot \hat{k}=1$

$\bullet$ $\bar{a}\cdot \bar{b}=\bar{b}\cdot \bar{a}$

$\bullet$ $\bar{a}\cdot \bar{o}=0$

$\bullet$ $\bar{a}\cdot \bar{b}=$(acos$\theta$)b = (projection of $\bar{a}on\bar{\mathbf{b}}$) b = (projection of $\bar{b}on\bar{a}$) a

$\bullet$$\bar{a}\cdot (\bar{b}+\bar{c})=\bar{a}\cdot \bar{b}+\bar{a}\cdot \bar{c}$(Distributive law)

$\bullet$ $(l\bar{a})\cdot (m\bar{b})=lm(\bar{a}\cdot \bar{b})$

$\bullet$ $\bar{a}\cdot \bar{b}=0$ $\iff$ $\bar{a},\bar{b}$are perpendicular to each other $\Rightarrow$ $\hat{i}\cdot \hat{j}=\hat{j}\cdot \hat{k}=\hat{k}\cdot \hat{i}=0$

$\bullet$ ${(\bar{a}\pm \bar{b})}^2=(\bar{a}\pm \bar{b})\cdot (\bar{a}\pm \bar{b})$ = ${\bar{a}}^2+{\bar{b}}^2\pm 2\bar{a}\cdot \bar{b}$

$\bullet$ $(\bar{a}+\bar{b})\cdot (\bar{a}-\bar{b})={\bar{a}}^2-{\bar{b}}^2=a^2-b^2$

$\bullet$ If $\bar{a}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k}and\bar{b}=b_1\hat{i}+b_2\hat{j}+b_3\hat{k}$ then$\bar{a}\cdot \bar{b}=a_1{b}_1+a_2{b}_2+a_3{b}_3$

$\bullet$ If $\bar{a},\bar{b}$ are non-zero, the angle between them is given by $\cos \theta =\frac{\bar{a}\cdot \bar{b}}{|\bar{a}||\bar{b}|}$

= $\frac{a_1{b}_1+a_2{b}_2+a_3{b}_3}{\sqrt{a_1^2+a_2^2+a_3^2}\sqrt{b_1^2+b_2^2+b_3^2}}$

Illustration : Prove by vector method that (a₁b₁ + a₂b₂ + a₃b₃)² $⩽$ (a₁²+ a₂² + a₃²) (b₁² + b₂² + b₃²).

Solution: Let $\vec{a}$ = a_1$\hat{i}$ + a₂ $\hat{j}$ + a_3$\hat{k}$ and $\vec{b}$ = $b_1\hat{i}+b_2\hat{j}+b_3\hat{k}$.

Now $\vec{a}$× $\vec{b}$ = a₁ b₁ + a₂b₂ + a₃b₃ , also $\vec{a}$× $\vec{b}$= |$\vec{a}$| |$\vec{b}$| cos$\theta$ $⩽$ |$\vec{a}$| |$\vec{b}$|.

$\Rightarrow$ ($\vec{a}$× $\vec{b}$)² $⩽$ |$\vec{a}$|² |$\vec{b}$|^{2 $\Rightarrow$} (a₁b₁ + a₂b₂ + a₃b₃)² $⩽$ $\left(a_1^2+a_2^2+a_3^2\right)\left(b_1^2+b_2^2+b_3^2\right)$