Class 11th

The vector product is also known as the cross product of two vectors. This results in a new vector. For all Class 11 students in Physics, one key concept of the cross product is that this resultant vector is perpendicular to the plane containing the original two vectors. The magnitude of this new vector is calculated by multiplying the magnitudes of the two initial vectors and the sine of the angle between them. The direction of the resulting vector is determined by the right-hand rule.

The product of two equal vectors depends on what type of multiplication we are using. 

1. Dot Product: The dot product of two equal vectors is the square of the magnitude of the vector. This is because the angle between two equal vectors is 0°, and the cosine of 0° is 1.

2. Cross Product: The vector product or cross product of two equal vectors is a zero vector (or null vector). This is because the angle between the vectors is zero, and the sine of 0° is 0.

The dot product of a vector with itself is the magnitude's square. When we multiply a vector by itself using the dot product, the angle between them is always zero. That's because the cosine of a 0° angle is 1. The dot product simplifies to the product of the magnitudes of the two vectors. And, in this case, it is the magnitude of the vector that is multiplied by itself. This operation results in a scalar quantity.

Motion of centre of mass has various applications in the real world. It is used for designing stable structures and vehicles, where low centre of mass make them more stable and faster. Even the motion of centre of mass applies to sportspeople, where athletes manipulate their centre of mass to optimise their performance. Apart from that, motion of centre of mass is required in animation and computer graphics to create realistic animations. 

The motion of individual particles relative to the centre of mass shows us how the particles move with respect to this central point. You can consider a frame of reference that moves along with the centre of mass. In this frame, the total momentum of all the particles in the system is nil or zero. 

The centre of mass moves as if all mass of the object is concentrated at one point while all external forces act on it. The motion of the centre of mass follows Newton's Second Law of Motion. That means that you can multiply the total mass of the system with the acceleration, which equals the net total force. The particles within the system have cancelled out internal forces, and do not affect the motion of the centre of mass.  

 $\frac{{\left(x+1\right)}^2}{\lambda +5}=\frac{{\left(y+1\right)}^2}{\frac{\lambda +5}{4}}=1$

length of latus rectum = $\frac{2b^2}{a}=\frac{2\left(\frac{\lambda +5}{4}\right)}{\sqrt[]{5+\lambda }}=4$

$\sqrt[]{\lambda +5}=8\Rightarrow \lambda =59$

Major axis = $2\sqrt[]{\lambda +5}=16$

$e_H=\sqrt[]{1+\frac{64}{49}}=\frac{\sqrt[]{113}}{7}$

$\Rightarrow e_H.e_E=\frac{1}{2}$

$\Rightarrow \frac{113}{49}.\frac{\left(64-a^2\right)}{64}=\frac{1}{4}\Rightarrow a^2-64=\frac{32^2}{113}$

$\mathcal{l}=\frac{2a^2}{b}=2\left(64+\frac{32^2}{113}\right).\frac{1}{8}$

$113\mathcal{l}=1552$

$\overline{x}=\frac{{\displaystyle \sum _{i=1}^{10}{x}_i}}{10}=15\frac{{\displaystyle \sum _{i=1}^{10}{x_i}^2}}{10}-{\left(\overline{x}\right)}^2=15$

$\Rightarrow \Sigma x_i=150\Sigma {x_i}^2=2400$

Actual mean $\overline{x}=\frac{\Sigma x_i+15-25}{10}=\frac{140}{10}=14$

Actual variance = $\frac{\Sigma {x_i}^2+15^2-25^2}{10}-{\left(14\right)}^2$

$=\frac{2400-400}{10}-196$

$\begin{array}{l}{\sigma }^2=4\\ \sigma =2\end{array}$