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We will now discuss the cross product which we also call the vector product or the outer product. This is an example of the rotational dynamics; here we will rotate one vector along the other vector. For example, we are having a two dimensional coordinate system. Let's say I have a vector \(\vec a\) which is completely horizontal and is having only one component, I have another vector \(\vec b\) which is inclined so it is in two components. \(\theta \) is the angle between vector \(\vec a\) and \(\vec b\). In the cross product, we are having the rotational dynamics so we will rotate vector \(\vec a\) along vector \(\vec b\) or vector \(\vec b\) along vector \(\vec a\).
Let's first rotates vector \(\vec a\) along vector \(\vec b\), if I consider a triangle then what is the area of this triangle?
If the height of the triangle is \(h\) then the area of triangle is $$ \frac { 1 }{ 2 } bh=A $$which means half times base \(x\) height. Now, if I split the parallelogram into two triangles, then I can write $$ \vec { a } \times \vec { b } =ab\sin { \theta \hat { n } } $$Here the base is \(a\) and the height is \(b\sin { \theta }\) $$ \vec { b } \times \vec { a } =ba\sin { \theta \hat { n } } $$ I represent the direction that is coming out of the board by a dot “.” and into the board direction is denoted by a cross “x”. So when we rotate \(\vec a\) along \(\vec b\) then we will have the dot direction here $$ \vec { a } \times \vec { b } =ab\sin { \theta \hat { . } } $$ And when we will rotate \(\vec b\) along \(\vec a\), then $$ \vec { b } \times \vec { a } =ba\sin { \theta \hat { \times } } $$ This is why we use \(\sin { \theta } \) in the cross product. So it is better instead of writing it this way, I write it as $$ \vec { a } \times \vec { b } =ab\sin { \theta \hat { n } } $$ $$ \vec { b } \times \vec { a } =ba\sin { \theta \left( -\hat { n } \right) } $$ Cross product is not a commutative relation $$ \vec { a } \times \vec { b } = - \vec { b } \times \vec { a } $$ Now, we will consider some limitations of \(\theta \), like dot product it's having its range $$0\le \theta \le { 180 }^{ o }$$ If \(\theta = { 0 }^{ o }\), means both the vectors are parallel to each other then we cannot rotate one vector around the other vector and the rotation will give us 0. $$ \vec { a } \times \vec { b } =0 $$ If \(\theta = { 45 }^{ o }\) $$ \vec { a } \times \vec { b } =\frac { ab }{ \sqrt { 2 } } $$ If \(\theta = { 60 }^{ o }\) $$ \vec { a } \times \vec { b } =\frac { ab }{ 2 } $$ If \(\theta = { 90 }^{ o }\) $$ \vec { a } \times \vec { b } = ab $$ This will give me the maximum area or the area of a square. Now I can consider this with an example of torque $$ \tau =\vec { r } \times \vec { F } =rF\sin { \theta \hat { n } } $$ \(r\) is the moment arm, \(F\) is the force and \(\tau \) is the rotational force. Consider a door; if my force will be exactly perpendicular to this door then I will be able to turn this door to the maximum. $$ 0<\theta \le { 90 }^{ o } $$ Similarly, we can take the example of angular momentum $$ \vec { L } =\vec { r } \times \vec { p } $$ where \(p\) is the linear momentum. We can take other examples as well. In component form, for example $$ \vec { a } =2\hat { x } +3\hat { y } +4\hat { z } $$ $$ \vec { b } =3\hat { x } +2\hat { y } +\hat { z } $$ What will be \(\vec a \times \vec b\)? $$ \vec { a } \times \vec { b } =\begin{vmatrix} \hat { x } & \hat { y } & \hat { z } \\ 2 & 3 & 4 \\ 3 & 2 & 1 \end{vmatrix} $$The inner components are giving zero and unlike the dot product, while, the outer components actually give us the result that's why it is also called outer product.
Summary:
• The cross product gives us the rotational dynamics while the dot product gives us the translational dynamics.
• When we multiply as a dot product, then \(\cos { \theta } \) is not coming by our choice but \(\cos { \theta } \) was actually due to the component. Similarly, \(\sin { \theta } \) here is not coming due to our choice but actually it is coming due to the height of vector.
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