Class 11th

Path length is a scalar quantity. It tells about the total distance covered, even though it depends on the path taken. Displacement is a vector quantity.  It tells that it's a straight-line change in position from the initial to the final point. Now, that is path independent. 

To find the direction in vector subtraction, let's consider (A – B).  We have to use vector addition by rewriting it as A + (–B). Then, this negative vector (–B) will have the same magnitude as B. Only that it will point in the opposite direction. Then we will use the head-to-tail method. Following that, we will place the tail of (–B) at the head of A. The resultant vector from the tail of A to the head of –B will give us both the magnitude and direction of A – B.

The graphical method using the head-to-tail or parallelogram laws only helps in visualising vectors and their resultants. But, it has limited accuracy. Because they cannot be precise when you consider the scale and angles. That's why it is important to use vector addition using the analytical method. That involves combining vector components. The graphical approach is primarily for conceptual understanding.

Scalar quantities only have magnitude, which makes sense to combine using ordinary algebra. But vector quantities have both magnitude and direction. Due to this directional aspect, vectors must obey special rules of vector algebra. Vectors have to specifically follow the triangle law or the parallelogram law of addition to be represented in the graph format. These graphical methods account for both magnitude and direction. This makes sure that the resultant vector accurately reflects the combined effect of the individual vectors. If we apply ordinary algebra, we won't be able to know the directional information.

$e=\frac{5}{4}$

$b^2=a^2\left(e^2-1\right)$

$\Rightarrow b=\frac{3a}{4}$

$p\left(\frac{8}{\sqrt[]{5}},\frac{12}{5}\right)$

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

$\Rightarrow \frac{64}{5a^2}-\frac{144}{25b^2}=1$

$\Rightarrow \frac{\sqrt[]{5}x}{3}+\frac{5}{8}y=\frac{8}{3}+\frac{3}{2}=\frac{25}{6}$

$a\in \left\{1,2,3,....,9\right\}$  

  $b\in \left\{0,1,2,...,9\right\}$              

9 * 9 = 81

81 + 81 + 81 = 243

$mean=\frac{n\left(n+1\right)/2}{n}$           

  $=\frac{n+1}{2}$            For n = 2k – 1, k   $\in N$

mean = k

mean deviation = avg. of deviations

$=\frac{2\left(\stackrel{0+}{1}+2+...\left(k-1\right)\right)}{2k-1}=\frac{\left(k-1\right)k}{2k-1}$

$=\frac{5\left(2k\right)}{k-1}\Rightarrow k=11\Rightarrow n=2k-1=21$            

               

${=}^{10}{C}_r.2^{10-r}.3^r.x^{30-3r-r}$  

  $For{x}^{30-4r}\to x^{\left(even\right)}$              

r = 0, 1, 2, 3, 4, 5, 6, 7

$r\in \left\{0,1,2,......,10\right\}$          

Sum of all such coefficient

$A=\left[\begin{array}{cc}\hfill 2\hfill & \hfill -2\hfill \\ \hfill 1\hfill & \hfill -1\hfill \end{array}\right]$ $B^2=\left[\begin{array}{cc}\hfill -1\hfill & \hfill 2\hfill \\ \hfill -1\hfill & \hfill 2\hfill \end{array}\right]\left[\begin{array}{cc}\hfill -1\hfill & \hfill 2\hfill \\ \hfill -1\hfill & \hfill 2\hfill \end{array}\right]$                  

$B=\left[\begin{array}{cc}\hfill -1\hfill & \hfill 2\hfill \\ \hfill -1\hfill & \hfill 2\hfill \end{array}\right]$ $=\left[\begin{array}{cc}\hfill -1\hfill & \hfill 2\hfill \\ \hfill -1\hfill & \hfill 2\hfill \end{array}\right]=B$

$A^2=\left[\begin{array}{cc}\hfill 2\hfill & \hfill -2\hfill \\ \hfill 1\hfill & \hfill -1\hfill \end{array}\right]\left[\begin{array}{cc}\hfill 2\hfill & \hfill -2\hfill \\ \hfill 1\hfill & \hfill -1\hfill \end{array}\right]$ $\Rightarrow .....B^3=B^2=B\Rightarrow B^n=B$

$=\left[\begin{array}{cc}\hfill 2\hfill & \hfill -2\hfill \\ \hfill 1\hfill & \hfill -1\hfill \end{array}\right]=A$

$\Rightarrow ...=A^3=A^2=A\Rightarrow A^n=A$

$n{A}^n+m{B}^m=l\Rightarrow nA+mB=l$

$\left[\begin{array}{cc}\hfill 2n\hfill & \hfill -2n\hfill \\ \hfill n\hfill & \hfill -n\hfill \end{array}\right]+\left[\begin{array}{cc}\hfill -m\hfill & \hfill 2m\hfill \\ \hfill -m\hfill & \hfill 2m\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$

$\Rightarrow 2n-m=1$  n – m = 0            ->n = m = 1

-2n + 2m = 0     -n + 2m = 1