Vectors
1.0 Introduction
2.0 Types of Vectors
3.0 Addition of Vectors
4.0 Components of a Vector
5.0 Vector Joining Two Points
6.0 Projection of a Vector on a Line
7.0 Section Formula
8.0 Products of a Vector
9.0 Lami's Theorem
10.0 Linear Combination of Vectors
11.0 Linearly Dependent and Independent Vectors
12.0 Scalar Triple Product
13.0 Vector Triple Product
8.2 Vector or Cross Product of two Vectors
It is similar to the right hand thumb used in the physics. figure
In the three dimensional right handed rectangular coordinate system, where positive $x$-axis is rotated counterclockwise into the positive $y$-axis, the right handed screw would be in the direction of positive $z$ axis.
Cross product of two non zero vectors $\overrightarrow a $ and $\overrightarrow b $ is denoted by $\overrightarrow a \times \overrightarrow b $ and is given as $$\overrightarrow a \times \overrightarrow b = \left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|sin\theta \widehat n$$where $\theta $ is the angle between $\overrightarrow a $ and $\overrightarrow b $ where $0 \leqslant \theta \leqslant \pi $ and $\widehat n$ is the unit vector perpendicular to both vectors such that $\overrightarrow a$ , $\overrightarrow b$ and $\widehat n$ form a right handed system, which means right hand system rotated from $\overrightarrow a to \overrightarrow b $ moves in the direction of $\widehat n$.
Note: If either of the vector is $0$, then $\theta $ is not defined and in this case $\overrightarrow a \times \overrightarrow b =0$.
Properties
Property 1: $\overrightarrow a \times \overrightarrow b $ is a vector.
Property 2: Let $\overrightarrow a $ and $\overrightarrow b $ be non zero vectors, then these vectors are said to be parallel (or collinear) if $\overrightarrow a \times \overrightarrow b=0 $
$$\overrightarrow a \times \overrightarrow b = 0 \Leftrightarrow \overrightarrow a \parallel \overrightarrow b $$
Property 3: $\overrightarrow a \times \overrightarrow a = 0$ , here $\theta = 0$ and $\overrightarrow a \times ( - \overrightarrow a ) = 0$ , here $\theta = \pi $
Property 4: If $\theta = \frac{\pi }{2}$ then $$\overrightarrow a \times \overrightarrow b = \left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|$$
Property 5: For mutually perpendicular vectors $\widehat i$, $\widehat j$ and $\widehat k$, we have $$\begin{equation} \begin{aligned} \widehat i \times \widehat i = \widehat j \times \widehat j = \widehat k \times \widehat k = 0; \\ \widehat i \times \widehat j = \widehat k,\widehat j \times \widehat k = \widehat i,\widehat k \times \widehat i = \widehat j \\\end{aligned} \end{equation} $$
The direction shown in the diagram must followed to get the correct answer. If the direction is changed, then negative sign is added as shown in the equation below. $$\hat j \times \widehat i = - \hat k,\;\hat k \times \hat j = - \hat i,\;\hat i \times \hat k = - \hat j$$
Property 6: In case of vector product , angle $\theta $ between two vectors is given as $$sin\theta = \frac{{\overrightarrow a \times \overrightarrow b }}{{\left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|}}$$
Property 7: Vector product is not commutative. Instead it is $\overrightarrow a \times \overrightarrow b = - \overrightarrow b \times \overrightarrow a $
Property 8: In case of $\vec a \times \vec b = \left| {\vec a} \right|\left| {\vec b} \right|sin\theta \,\hat n$ , $\overrightarrow a$ ,$\overrightarrow b$ and $\widehat n$ form a right handed system where $\theta $ is traversed from $\overrightarrow a to \overrightarrow b $ whereas in this case of $\vec b \times \vec a = \left| {\vec a} \right|\left| {\vec b} \right|sin\theta \,\widehat {{n_1}}$, where $\overrightarrow b ,\overrightarrow a and \widehat {{n_1}}$ form right handed system i.e., $\theta $ is traversed from $\overrightarrow b to \overrightarrow a $. Thus, if we assume that $\overrightarrow a $ and $\overrightarrow b $ to lie in the plane of paper then $\widehat n$ and $\widehat {{n_1}}$ both will be perpendicular to the plane of paper. But $\widehat n$ is directed above the paper while $\widehat {{n_1}}$ is directed below the paper such that $$\widehat {{n_1}} = - \widehat n$$Hence, $$\begin{equation} \begin{aligned} \vec a \times \vec b = \left| {\vec a} \right|\left| {\vec b} \right|sin\theta \,\hat n \\ {\text{ }} = - \left| {\vec a} \right|\left| {\vec b} \right|sin\theta \,\widehat {{n_1}} = - \vec b \times \vec a \\\end{aligned} \end{equation} $$
Property 9: From above property, we have $$\widehat j \times \widehat i = - \widehat k;\widehat k \times \widehat j = - \widehat i;\widehat i \times \widehat k = - \widehat j$$
Property 10: If the $\overrightarrow a $ and $\overrightarrow b $ represents the adjacent sides of a triangle then area of triangle is given by $$\frac{1}{2}\left| {\overrightarrow a \times \overrightarrow b } \right|$$Since area of triangle $ABC$ is given by $\frac{1}{2}AB.CD$.
We know that $$\begin{equation} \begin{aligned} AB = \left| {\overrightarrow b } \right| \\ CD = \left| {\overrightarrow a } \right|sin\theta \\\end{aligned} \end{equation} $$
Therefore, area of triangle is given as $$\frac{1}{2}\left| {\overrightarrow b } \right|\left| {\overrightarrow a } \right|sin\theta = \frac{1}{2}\left| {\overrightarrow a \times \overrightarrow b } \right|$$
Property 11: If the $\overrightarrow a $ and $\overrightarrow b $ represents the adjacent sides of a parallelogram then area of parallelogram is given by $$\left| {\overrightarrow a \times \overrightarrow b } \right|$$Area of parallelogram $ABCD$ =$AB.DE$ but $$\begin{equation} \begin{aligned} AB = \left| {\overrightarrow b } \right| \\ DE = \left| {\overrightarrow a } \right|sin\theta \\\end{aligned} \end{equation} $$
Area of parallelogram $ABCD$ is $$=\left| {\overrightarrow b } \right|\left| {\overrightarrow a } \right|sin\theta = \left| {\overrightarrow a \times \overrightarrow b } \right|$$
Property 12: Vector or cross product of two vectors $\overrightarrow a $ and $\overrightarrow b $ can be given as
\[\left( {\begin{array}{c} {\widehat i}&{\widehat j}&{{{\widehat k}_{}}} \\ {{a_1}}&{{a_2}}&{{a_3}} \\ {{b_1}}&{{b_2}}&{{b_3}} \end{array}} \right)\]
Proof: $$\begin{equation} \begin{aligned} \vec a \times \vec b = {a_1}{b_1}(\hat i \times \hat i) + {a_1}{b_2}(\hat i \times \hat j) + {a_1}{b_3}(\hat i \times \hat k) + {a_2}{b_1}(\hat j \times \hat i) + \\ {\text{ }}{a_2}{b_2}(\hat j \times \hat j) + {a_2}{b_3}(\hat j \times \hat k) + {a_3}{b_1}(\hat k \times \hat i) + {a_3}{b_2}(\hat k \times \hat j) + {a_3}{b_3}(\hat k \times \hat k) \\\end{aligned} \end{equation} $$
$$\overrightarrow a \times \overrightarrow b = {a_1}{b_2}(\widehat i \times \widehat j) - {a_1}{b_3}(\widehat k \times \widehat i) - {a_2}{b_1}(\widehat i \times \widehat j) + {a_2}{b_3}(\widehat j \times \widehat k) + {a_3}{b_1}(\widehat k \times \widehat i) - {a_3}{b_2}(\widehat j \times \widehat k)$$Using property $5$
$$\begin{equation} \begin{aligned} \overrightarrow a \times \overrightarrow b = {a_1}{b_2}\widehat k - {a_1}{b_3}\widehat j - {a_2}{b_1}\widehat k + {a_2}{b_3}\widehat i + {a_3}{b_1}\widehat j - {a_3}{b_2}\widehat i \\ \overrightarrow a \times \overrightarrow b = ({a_2}{b_3} - {a_3}{b_2})\widehat i + ({a_3}{b_1} - {a_1}{b_3})\widehat j + ({a_1}{b_2} - {a_2}{b_1})\widehat k \\\end{aligned} \end{equation} $$
Property 13: Vector product is distributive over addition: Let $\overrightarrow a ,\overrightarrow b\ and\ \overrightarrow c $ in the component form as $\overrightarrow a = ({a_1}\widehat i + {a_2}\widehat j + {a_3}\widehat k)$ and $\overrightarrow b = ({b_1}\widehat i + {b_2}\widehat j + {b_3}\widehat k)$ and $\lambda $ be any scalar, then
$$\begin{equation} \begin{aligned} \overrightarrow a \times (\overrightarrow b + \overrightarrow c ) = (\overrightarrow a \times \overrightarrow b ) + (\overrightarrow a \times \overrightarrow c ) \\ \lambda (\overrightarrow a \times \overrightarrow b ) = (\lambda \overrightarrow a ) \times \overrightarrow b = \overrightarrow a \times (\lambda \overrightarrow b ) \\\end{aligned} \end{equation} $$
Property 14: Vector addition is associative
$$(\overrightarrow a + \overrightarrow b ) \times (\overrightarrow c + \overrightarrow d ) = (\overrightarrow a \times \overrightarrow c ) + (\overrightarrow a \times \overrightarrow d ) + (\overrightarrow b \times \overrightarrow c ) + (\overrightarrow b \times \overrightarrow d )$$
Proof: Let vectors $\overrightarrow a = {a_1}\widehat i + {a_2}\widehat j + {a_3}\widehat k$ ,$\overrightarrow b = {b_1}\widehat i + {b_2}\widehat j + {b_3}\widehat k$, $\overrightarrow c = {c_1}\widehat i + {c_2}\widehat j + {c_3}\widehat k$ , and $\overrightarrow d = {d_1}\widehat i + {d_2}\widehat j + {d_3}\widehat k$
So considering LHS, we have $$(\overrightarrow a + \overrightarrow b ) \times (\overrightarrow c + \overrightarrow d )$$$$\begin{equation} \begin{aligned} \{ ({a_1}\widehat i + {a_2}\widehat j + {a_3}\widehat k) + ({b_1}\widehat i + {b_2}\widehat j + {b_3}\widehat k)\} \times \{ ({c_1}\widehat i + {c_2}\widehat j + {c_3}\widehat k) + ({d_1}\widehat i + {d_2}\widehat j + {d_3}\widehat k)\} \\ = \{ ({a_1} + {b_1})\widehat i + ({a_2} + {b_2})\widehat j + ({a_3} + {b_3})\widehat k\} \times \{ ({c_1} + {d_1})\widehat i + ({c_2} + {d_2})\widehat j + ({c_3} + {d_3})\widehat k\} \\\end{aligned} \end{equation} $$
Now using matrix for cross product, we have
\[\left( {\begin{array}{c} {\widehat i}&{\widehat j}&{{{\widehat k}_{}}} \\ {({a_1} + {b_1})}&{({a_2} + {b_2})}&{({a_3} + {b_3})} \\ {({c_1} + {d_1})}&{({c_2} + {d_2})}&{({c_3} + {d_3})} \end{array}} \right)\]
$$\begin{equation} \begin{aligned} = \{ ({a_2} + {b_2})({c_3} + {d_3}) - ({c_2} + {d_2})({a_3} + {b_3})\} \hat i \\ - \{ ({a_1} + {b_1})({c_3} + {d_3}) - ({c_1} + {d_1})({a_3} + {b_3})\} \hat j \\ + \{ ({a_1} + {b_1})({c_2} + {d_2}) - ({c_1} + {d_1})({a_2} + {b_2})\} \hat k \\\end{aligned} \end{equation} $$$$\begin{equation} \begin{aligned} = \{ ({a_2}{c_3} - {c_2}{a_3}) + ({a_2}{d_3} - {d_2}{a_3}) + ({b_2}{c_3} - {c_2}{b_3}) + ({b_2}{d_3} - {d_2}{b_3})\} \hat i \\ - \{ ({a_1}{c_3} - {c_1}{a_3}) + ({a_1}{d_3} - {d_1}{a_3}) + ({b_1}{c_3} - {c_1}{b_3}) + ({b_1}{d_3} - {d_1}{b_3})\} \hat j \\ + \{ ({a_1}{c_2} - {c_1}{a_2}) + ({a_1}{d_2} - {d_1}{a_2}) + ({b_1}{c_2} - {c_1}{b_2}) + ({b_1}{d_2} - {d_1}{b_2})\} \hat k \\\end{aligned} \end{equation} $$
RHS: \[\begin{gathered}(\vec a \times \vec c) + (\vec a \times \vec d) + (\vec b \times \vec c) + (\vec b \times \vec d) = \{ ({a_1}\hat i + {a_2}\hat j + {a_3}\hat k) \times ({c_1}\hat i + {c_2}\hat j + {c_3}\hat k)\} + \hspace{1em} \\{\text{ }}\{ ({a_1}\hat i + {a_2}\hat j + {a_3}\hat k) \times ({d_1}\hat i + {d_2}\hat j + {d_3}\hat k)\} + \hspace{1em} \\{\text{ }}\{ ({b_1}\hat i + {b_2}\hat j + {b_3}\hat k) \times ({c_1}\hat i + {c_2}\hat j + {c_3}\hat k)\} + \hspace{1em} \\{\text{ }}\{ ({b_1}\hat i + {b_2}\hat j + {b_3}\hat k) \times ({d_1}\hat i + {d_2}\hat j + {d_3}\hat k)\} \hspace{1em} \\ \end{gathered} \]
\[ = \left( {\begin{array}{c} {\widehat i}&{\widehat j}&{{{\widehat k}_{}}} \\{{a_1}}&{{a_2}}&{{a_3}} \\{{c_1}}&{{c_2}}&{{c_3}} \end{array}} \right) + \left( {\begin{array}{c} {\widehat i}&{\widehat j}&{{{\widehat k}_{}}} \\{{a_1}}&{{a_2}}&{{a_3}} \\ {{d_1}}&{{d_2}}&{{d_3}}\end{array}} \right) + \left( {\begin{array}{c} {\widehat i}&{\widehat j}&{{{\widehat k}_{}}} \\ {{b_1}}&{{b_2}}&{{b_3}} \\ {{c_1}}&{{c_2}}&{{c_3}} \end{array}} \right) + \left( {\begin{array}{c}{\widehat i}&{\widehat j}&{{{\widehat k}_{}}} \\{{b_1}}&{{b_2}}&{{b_3}} \\{{d_1}}&{{d_2}}&{{d_3}} \end{array}} \right)\]
$$\begin{equation} \begin{aligned} = \{ ({a_2}{c_3} - {c_2}{a_3})\hat i - ({a_1}{c_3} - {c_1}{a_3})\hat j + ({a_1}{c_2} - {c_1}{a_2})\hat k\} \\ + \{ ({a_2}{d_3} - {d_2}{a_3})\hat i - ({a_1}{d_3} - {d_1}{a_3})\hat j + ({a_1}{d_2} - {d_1}{a_2})\hat k\} \\ + \{ ({b_2}{c_3} - {c_2}{b_3})\hat i - ({b_1}{c_3} - {c_1}{b_3})\hat j + ({b_1}{c_2} - {c_1}{b_2})\hat k\} \\ + \{ ({b_2}{d_3} - {d_2}{b_3})\hat i - ({b_1}{d_3} - {d_1}{b_3})\hat j + ({b_1}{d_2} - {d_1}{b_2})\hat k\} \\\end{aligned} \end{equation} $$$$\begin{equation} \begin{aligned} = \{ ({a_2}{c_3} - {c_2}{a_3}) + ({a_2}{d_3} - {d_2}{a_3}) + ({b_2}{c_3} - {c_2}{b_3}) + ({b_2}{d_3} - {d_2}{b_3})\} \hat i \\ - \{ ({a_1}{c_3} - {c_1}{a_3}) + ({a_1}{d_3} - {d_1}{a_3}) + ({b_1}{c_3} - {c_1}{b_3}) + ({b_1}{d_3} - {d_1}{b_3})\} \hat j \\ + \{ ({a_1}{c_2} - {c_1}{a_2}) + ({a_1}{d_2} - {d_1}{a_2}) + ({b_1}{c_2} - {c_1}{b_2}) + ({b_1}{d_2} - {d_1}{b_2})\} \hat k \\\end{aligned} \end{equation} $$
