Scalar Product of Two Vectors

When two vectors are multiplied in such a way that their product is a scalar quantity then it is called scalar product or dot product of two vectors.

Let $\overrightarrow{a}=(a_1,a_2)$ and $\overrightarrow{b}=(b_1,b_2)$ be any two plane vectors, then the scalar product of two vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ denoted by $\overrightarrow{a}\cdot\overrightarrow{b}$ (read $\overrightarrow{a}$ dot $\overrightarrow{b}$) is defined by \[\overrightarrow{a}\cdot\overrightarrow{b}=a_1b_1+a_2b_2\]

Again, if $\overrightarrow{a}=(a_1,a_2,a_3)$ and $\overrightarrow{b}=(b_1,b_2,b_3)$ are two space vectors then the scalar product of two vectors is defined by \[\overrightarrow{a}\cdot\overrightarrow{b} =a_1b_1+a_2b_2+a_3b_3\]

The scalar product of two vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ is also equal to the product of the magnitudes of $\overrightarrow{a}$ and $\overrightarrow{b}$ and the cosine of the angle between them. If $\theta$ is the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$, then \[\overrightarrow{a}\cdot\overrightarrow{b}=|\overrightarrow{a}||\overrightarrow{b}|\cos\theta =ab\cos \theta\]

Here’s a free tool you can use to calculate dot product between two vectors.

Angle between Two Vectors

Let $O$ be the origin and $OX$ and $OY$ be the two mutually perpendicular straight lines representing x-axis and y-axis respectively. Let $\overrightarrow{OA}$$=\overrightarrow{a}$$=(a_1,a_2)$ and $\overrightarrow{OB}$$=\overrightarrow{b}$$=(b_1,b_2)$ be the two vectors. Again, let $\angle XOA= \beta$ and $\angle XOB=\alpha$ so that $\angle AOB=\alpha-\beta=\theta$ (say).

From $A$ and $B$, draw $AM$ and $BN$ perpendiculars to $OX$. Then, in $\Delta OMA$, \[\cos\beta =\frac{OM}{OA}\] \[OM=OA\cos\beta\] \[\therefore a_1=a\cos\beta\] and, \[\sin\beta=\frac{MA}{OA}\] \[MA=OA\sin\beta\] \[\therefore a_2=a\sin\beta\]

Similarly, in $\Delta ONB$, \[b_1=b\cos\alpha\;\;\text{and}\;\;b_2=b\sin\alpha\] where $a$ and $b$ are the magnitudes of $\overrightarrow{a}$ and $\overrightarrow{b}$ respectively.

Now, \[\overrightarrow{a}\cdot\overrightarrow{b}=a_1b_1+a_2b_2\] \[=a\cos\beta\:b\cos\alpha+a\sin \beta\:b\sin\alpha\] \[=ab(\cos\alpha\cos\beta+\sin\alpha\sin\beta)\] \[=ab\cos(\alpha-\beta)\] \[=ab\cos\theta\]

Hence, the angle between two vectors is given by, \[\cos\theta=\frac{\overrightarrow{a}\cdot \overrightarrow{b}}{ab}\] Here, \[\overrightarrow{a}\cdot\overrightarrow{b}=a_1b_1+a_2b_2\] \[|\overrightarrow{a}|=a=\sqrt{a_1^2+a_2^2}\] \[|\overrightarrow{b}|=b=\sqrt{b_1^2+b_2^2}\] \[\therefore\cos\theta=\frac{\overrightarrow{a}\cdot \overrightarrow{b}}{ab}=\frac{a_1b_1+a_2b_2} {\sqrt{a_1^2+a_2^2}\sqrt{b_1^2+b_2^2}}\]

If $\overrightarrow{a}=(a_1,a_2,a_3)$ and $\overrightarrow{b}=(b_1,b_2,b_3)$ be two space vectors, then, \[\cos\theta=\frac{\overrightarrow{a}\cdot \overrightarrow{b}}{ab}=\frac{a_1b_1+a_2b_2+a_3b_3} {\sqrt{a_1^2+a_2^2+a_3^2}\sqrt{b_1^2+b_2^2+b_3^2}}\]

Parallelism and Perpendicularity

Let $\overrightarrow{a}=(a_1,a_2,a_3)$ and $\overrightarrow{b}=(b_1,b_2,b_3)$ be two space vectors.

When two vectors are parallel, then $\theta=0°$. \[\therefore\overrightarrow{a}\cdot\overrightarrow{b} =ab\cos 0°=ab\]

It means that the dot product of two parallel vectors is equal to product of their magnitudes.

When two vectors are perpendicular, then $\theta=90°$. \[\therefore\overrightarrow{a}\cdot \overrightarrow{b}=(a_1,a_2,a_3)\cdot(b_1,b_2,b_3)\] \[=a_1b_1+a_2b_2+a_3b_3\] \[=ab\cos 90° =0\]

Thus, if two vectors are perpendicular to each other, their scalar product must be zero.

Geometrical Interpretation of Scalar Product of Two Vectors

Let the two vectors be $\overrightarrow{a}$ and $\overrightarrow{b}$ be represented by $\overrightarrow{OA}$ and $\overrightarrow{OB}$ respectively. Let $\theta$ be the angle between the vectors. \[\therefore\overrightarrow{OA}=\overrightarrow{a}\;\;\text{and}\;\;\overrightarrow{OB} =\overrightarrow{b}\]

From $A$ and $B$, draw $AD$ and $BE$ perpendiculars to $OB$ and $OA$ respectively.

Now, \[\overrightarrow{a}\cdot\overrightarrow{b}=ab\cos\theta=(OA)(OB\cos\theta)=OA×OE\] $\therefore\overrightarrow{a}\cdot\overrightarrow{b}=$ magnitude of $\overrightarrow{a}$ $×$ projection of $\overrightarrow{b}$ on $\overrightarrow{a}$.

Similarly, $\overrightarrow{a}\cdot\overrightarrow{b}=$ magnitude of $\overrightarrow{b}$ $×$ projection of $\overrightarrow{a}$ on $\overrightarrow{b}$.

Hence, the scalar product of two vectors is defined as the product of the magnitudes of one of the vectors and the projection of the second vector on the first.

OR, \[\overrightarrow{a}\cdot\overrightarrow{b}=ab\cos\theta=a(b\cos\theta)\] $\therefore \overrightarrow{a}\cdot\overrightarrow{b}=$ $a$ (magnitude of component vector of $\overrightarrow{b}$ in the direction of $\overrightarrow{a}$)

Similarly, $\overrightarrow{a}\cdot\overrightarrow{b}=$ $b$ (magnitude of component vector of $\overrightarrow{a}$ in the direction of $\overrightarrow{b}$)

Hence, the scalar product or dot product of two vectors is also defined as the product of the magnitude of one vector with the magnitude of the component of the other vector in the direction of first vector.

Length of a Vector as a Scalar Product

Let $\overrightarrow{a}=(a_1,a_2)$ be a plane vector, then \[\overrightarrow{a}\cdot\overrightarrow{b} =(a_1,a_2)\cdot(a_1,a_2)\] \[=a_1^2+a_2^2=a^2\]

Since $a_1^2+a_2^2>0$ for $a_1=a_2≠0$, so \[a=\sqrt{a_1^2+a_2^2}=\sqrt{\overrightarrow{a}\cdot \overrightarrow{a}}\]

Hence, the length of a vector $\overrightarrow{a}$ is the positive square root of the scalar product $\overrightarrow{a}\cdot\overrightarrow{a}$. The scalar product of a vector with itself is often written as the square of the vector.

Mutually Perpendicular Unit Vectors

The unit vectors $\overrightarrow{i}$ and $\overrightarrow{j}$ in a plane are defined by $(1,0)$ and $(0,1)$ respectively. Now, \[\overrightarrow{i}\cdot\overrightarrow{j}=(1,0)\cdot(0,1)=0+0=0= \overrightarrow{j}\cdot\overrightarrow{i}\]

and, \[\overrightarrow{i}\cdot\overrightarrow{i}=(1,0)\cdot(1,0)=1+0=1\] \[\text{Similarly,} \;\overrightarrow{j}\cdot \overrightarrow{j}=1\]

Again, the unit vectors $\overrightarrow{i}$, $\overrightarrow{j}$ and $\overrightarrow{k}$ in space are defined by $(1,0,0)$, $(0,1,0)$ and $(0,0,1)$ respectively.

Now, \[\overrightarrow{i}\cdot\overrightarrow{i}=(1,0,0)\cdot(1,0,0)=1+0+0=1\] \[\text{Similarly,} \;\overrightarrow{j}\cdot \overrightarrow{j}=\overrightarrow{k}\cdot \overrightarrow{k}=1\]

Again, \[\overrightarrow{i}\cdot\overrightarrow{j}=(1,0,0)\cdot(0,1,0)=0+0+0=0=\overrightarrow{j}\cdot \overrightarrow{i}\] Similarly, \[\overrightarrow{j}\cdot\overrightarrow{k}=\overrightarrow{k}\cdot \overrightarrow{j}=0\] and, \[\overrightarrow{k}\cdot\overrightarrow{i}=\overrightarrow{i}\cdot \overrightarrow{k}=0\]

The value of the dot product of two mutually perpendicular unit vectors is shown in the following table:

Properties of Scalar Product

Commutative Law

The scalar product of two vectors is commutative i.e. if $\overrightarrow{a}$ and $\overrightarrow{b}$ be the two vectors, then \[\overrightarrow{a}\cdot\overrightarrow{b}=\overrightarrow{b}\cdot \overrightarrow{a}\]

Proof: Let $\overrightarrow{a}=(a_1,a_2,a_3)$ and $\overrightarrow{b}=(b_1,b_2,b_3)$ be two space vectors. Then, \[\overrightarrow{a}\cdot\overrightarrow{b}=(a_1,a_2,a_3)\cdot(b_1,b_2,b_3)\] \[=a_1b_1+a_2b_2+a_3b_3\] \[=b_1a_1+b_2a_2+b_3a_3\] \[=\overrightarrow{b}\cdot\overrightarrow{a}\]

Associative Law

The scalar product of two vectors is associative i.e. if $\overrightarrow{a}$ and $\overrightarrow{b}$ be any two vectors and $m$, $n$ any scalars, then \[m\:\overrightarrow{a}\cdot n\:\overrightarrow{b}=mn (\overrightarrow{a}\cdot\overrightarrow{b})=mn\:\overrightarrow{a}\cdot\overrightarrow{b}=\overrightarrow{a}\cdot mn\:\overrightarrow{b}\]

Proof: Let $\overrightarrow{a}=(a_1,a_2,a_3)$ and $\overrightarrow{b}=(b_1,b_2,b_3)$ be two space vectors. Then,

\[m\:\overrightarrow{a}\cdot n\:\overrightarrow{b}=(ma_1,ma_2,ma_3)\cdot(nb_1,nb_2,nb_3)\] \[=mna_1b_1+mna_2b_2+mna_3b_3\] \[=mn(a_1b_1+a_2b_2+a_3b_3)\] \[=mn(\overrightarrow{a}\cdot \overrightarrow{b})\]

\[mn(\overrightarrow{a}\cdot \overrightarrow{b})=mn(a_1b_1+a_2b_2+a_3b_3)\] \[=(mna_1)b_1+(mna_2)b_2+(mna_3)b_3\] \[=mn\:\overrightarrow{a}\cdot\overrightarrow{b}\] Similarly, \[mn(\overrightarrow{a}\cdot \overrightarrow{b})=\overrightarrow{a}\cdot mn\:\overrightarrow{b}\]

\[\therefore m\:\overrightarrow{a}\cdot n\:\overrightarrow{b}=mn (\overrightarrow{a}\cdot \overrightarrow{b})=mn\:\overrightarrow{a}\cdot\overrightarrow{b}=\overrightarrow{a}\cdot mn\:\overrightarrow{b}\]

Distributive Law

If $\overrightarrow{a}$, $\overrightarrow{b}$ and $\overrightarrow{c}$ be any three vectors, then \[\overrightarrow{a}\cdot(\overrightarrow{b}+ \overrightarrow{c})=\overrightarrow{a}\cdot \overrightarrow{b}+\overrightarrow{a}\cdot \overrightarrow{c}\]

Proof: Let $\overrightarrow{a}=(a_1,a_2,a_3)$, $\overrightarrow{b}=(b_1,b_2,b_3)$ and $\overrightarrow{c}=(c_1,c_2,c_3)$ be any three space vectors. Then,

\[\overrightarrow{b}+\overrightarrow{c}=(b_1+c_1,b_2+c_2,b_3+c_3)\] \[\overrightarrow{a}\cdot \overrightarrow{b}=a_1b_1+a_2b_2+a_3b_3\] \[\overrightarrow{a}\cdot\overrightarrow{c}=a_1c_1 +a_2c_2+a_3c_3\]

Now, \[\overrightarrow{a}\cdot(\overrightarrow{b}+\overrightarrow{c})\] \[=(a_1,a_2,a_3)\cdot (b_1+c_1, b_2+c_2, b_3+c_3)\] \[=a_1(b_1+c_1)+a_2(b_2+c_2)+a_3(b_3+c_3)\] \[=a_1b_1+a_1c_1+a_2b_2+a_2c_2+a_3b_3+a_3c_3\] \[=(a_1b_1+a_2b_2+a_3b_3)+(a_1c_1+a_2c_2+a_3c_3)\] \[=\overrightarrow{a}\cdot\overrightarrow{b}+ \overrightarrow{a}\cdot \overrightarrow{c}\]

\[\therefore\overrightarrow{a}\cdot(\overrightarrow{b}+ \overrightarrow{c})=\overrightarrow{a}\cdot \overrightarrow{b}+\overrightarrow{a}\cdot \overrightarrow{c}\]

Examples of Scalar Product of Two Vectors

• Work done $(W)$ is defined as the dot product of force vector $(\overrightarrow{F})$ and displacement vector $(\overrightarrow{S})$. \[\therefore W= \overrightarrow{F}\cdot \overrightarrow{S}=FS\cos\theta\]
• The instantaneous power $(P)$ is defined as the dot product of force vector $(\overrightarrow{F})$ and instantaneous velocity $(\overrightarrow{v})$. \[\therefore P= \overrightarrow{F}\cdot \overrightarrow{v}=Fv\cos\theta\]
• Magnetic flux $(\phi)$ linked with a surface is defined as the dot product of magnetic induction vector $(\overrightarrow{B})$ and area vector $(\overrightarrow{A})$. \[\therefore\phi= \overrightarrow{B}\cdot \overrightarrow{A}=BA\cos\theta\]

Here, work done, power and magnetic flux are scalar quantities.

Next: Application of Dot Product to Plane Trigonometry