By Dr. Shahid Ali Yousafzai

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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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