Consider the lines $L_1$ and $L_2$ given by 
$L_1: \frac{x-1}{2}=\frac{y-3}{1}=\frac{z-2}{2} $  
$L_2: \frac{x-2}{1}=\frac{y-2}{2}=\frac{z-3}{3} $ 
A line $L_3$ having direction ratios $1,-1,-2$, intersects $L_1$ and $L_2$ at the points $P$ and $Q$ respectively Then the length of line segment $P Q$ is

The correct answer is (C) : \(2\sqrt6\)
Let P = (2λ+1,λ+3,2λ+2)
Let Q = \((\mu+2,2\mu+2,3\mu+3)\)
\(⇒\frac{2λ-\mu-1}{1}=\frac{λ-2\mu-1}{-1}\)
\(=\frac{2λ-3\mu-1}{-2}⇒λ=\mu=3\)
\(⇒P(7,6,8)\) and \(Q(5,8,12)\)
PQ = \(2\sqrt6\)
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is:
If some other quantity ‘y’ causes some change in a quantity of surely ‘x’, in view of the fact that an equation of the form y = f(x) gets consistently pleased, i.e, ‘y’ is a function of ‘x’ then the rate of change of ‘y’ related to ‘x’ is to be given by
\(\frac{\triangle y}{\triangle x}=\frac{y_2-y_1}{x_2-x_1}\)
This is also known to be as the Average Rate of Change.
Consider y = f(x) be a differentiable function (whose derivative exists at all points in the domain) in an interval x = (a,b).
Read More: Application of Derivatives