Physics > Basic Vectors > 10.0 Dot product of two vectors
Basic Vectors
1.0 Introduction
2.0 Representation of vector
3.0 Basic definition related with vectors
3.1 Unit vector
3.2 Negative of a vector
3.3 Modulus of a vector
3.4 Multiplication and division of vectors by scalars
4.0 Types of vectors
4.1 Equal vectors
4.2 Parallel vectors
4.3 Anti-parallel vectors
4.4 Collinear vectors
4.5 Coplanar vectors
4.6 Zero or null vectors
5.0 Angle between the vectors
6.0 Addition of vectors
6.1 Triangle law of vector addition
6.2 Parallelogram law of vector addition
6.3 Relation between triangle and parallelogram law of vector addition
7.0 Subtraction of vectors
8.0 Cartesian co-ordinate system
8.1 Unit vector in cartesian co-ordinate system
8.2 Position vector of a point
8.3 Displacement vector
9.0 Resolving vector into its components
10.0 Dot product of two vectors
10.1 Properties of dot product
10.2 Condition when two vectors are perpendicular
10.3 Angle between two vectors
10.4 Geometrical meaning of scalar product
10.5 Application of dot product
11.0 Cross product of two vectors
10.1 Properties of dot product
3.2 Negative of a vector
3.3 Modulus of a vector
3.4 Multiplication and division of vectors by scalars
4.2 Parallel vectors
4.3 Anti-parallel vectors
4.4 Collinear vectors
4.5 Coplanar vectors
4.6 Zero or null vectors
6.2 Parallelogram law of vector addition
6.3 Relation between triangle and parallelogram law of vector addition
8.2 Position vector of a point
8.3 Displacement vector
10.2 Condition when two vectors are perpendicular
10.3 Angle between two vectors
10.4 Geometrical meaning of scalar product
10.5 Application of dot product
1. $\overrightarrow A .\overrightarrow B = \left| {\overrightarrow A } \right|\left| {\overrightarrow B } \right|\cos \theta $ or $\overrightarrow A .\overrightarrow B = AB\cos \theta $
2. $\overrightarrow A .\overrightarrow B = \overrightarrow B .\overrightarrow A $ (Commutative law)
As, $$\overrightarrow A .\overrightarrow B = AB\cos \theta = BA\cos \theta = \overrightarrow B .\overrightarrow A $$
3. $\left| {\widehat i} \right| = \left| {\widehat j} \right| = \left| {\widehat k} \right| = 1$
$\widehat i$, $\widehat j$ & $\widehat k$ are three mutually perpendicular vectors.
As, $$\overrightarrow A .\,\overrightarrow B = \left| {\overrightarrow A } \right|\left| {\overrightarrow B } \right|\cos \theta $$ So, $$\widehat i.\widehat i = \left( 1 \right)\left( 1 \right)\cos 0^\circ $$$$\widehat i.\widehat i = 1$$ Similarly, $$\widehat j.\widehat j = 1$$$$\widehat k.\widehat k = 1$$ Also, $$\widehat j.\widehat i = \widehat i.\widehat j = \left( 1 \right)\left( 1 \right)\cos 90^\circ = 0$$$$\widehat i.\widehat k = \widehat k.\widehat i = \left( 1 \right)\left( 1 \right)\cos 90^\circ = 0$$$$\widehat j.\widehat k = \widehat k.\widehat j = \left( 1 \right)\left( 1 \right)\cos 90^\circ = 0$$
4. Consider vectors $\overrightarrow A $ and $\overrightarrow B$ are given as, $$\overrightarrow A = a\widehat i + b\widehat j + c\widehat k$$$$\overrightarrow B = p\widehat i + q\widehat j + r\widehat k$$$$\overrightarrow A .\overrightarrow B = ap + bq + cr$$
Note: In dot product simply multiply $\widehat i$ term of $\overrightarrow A $ with $\widehat i$ term of $\overrightarrow B $. Similarly for $\widehat j$ & $\widehat k$ and add all the three terms.
Proof:
$$\overrightarrow A .\overrightarrow B = \left( {a\widehat i + b\widehat j + c\widehat k} \right)\left( {p\widehat i + q\widehat j + r\widehat k} \right)$$$$\overrightarrow A .\overrightarrow B = ap\left( {\widehat i.\widehat i} \right) + aq\left( {\widehat i.\widehat j} \right) + ar\left( {\widehat i.\widehat k} \right) + bp\left( {\widehat j.\widehat i} \right) + bq\left( {\widehat j.\widehat j} \right) + br\left( {\widehat j.\widehat k} \right) + cp\left( {\widehat k.\widehat i} \right) + cq\left( {\widehat k.\widehat j} \right) + cr\left( {\widehat k.\widehat k} \right)$$
As, $$\widehat i.\widehat i = \widehat j.\widehat j = \widehat k.\widehat k = 1$$$$\widehat i.\widehat j = \widehat j.\widehat i = \widehat j.\widehat k = \widehat k.\widehat j = \widehat k.\widehat i = \widehat i.\widehat k = 0$$
So, $$\overrightarrow A .\overrightarrow B = ap + bq + cr$$
5. $\overrightarrow A .\overrightarrow B = \left| {\overrightarrow A } \right|\left| {\overrightarrow B } \right|\cos \theta $
If $\theta = 0^\circ $ (when vectors are parallel)
$$\overrightarrow A .\overrightarrow B = AB$$
If $\theta = 90^\circ $ (when vectors are perpendicular)$$\overrightarrow A .\overrightarrow B = AB\cos 90^\circ $$$$\overrightarrow A .\overrightarrow B = 0$$
If $\theta = 180^\circ $ (when vectors are antiparallel)
$$\overrightarrow A .\overrightarrow B = AB\cos 180^\circ $$$$\overrightarrow A .\overrightarrow B = - AB$$
