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
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
- $3 \sqrt{2}$
- 4
- $2 \sqrt{6}$
- $4 \sqrt{3}$
The Correct Option is C
Solution and Explanation
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 = \(2\sqrt6\)
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Concepts Used:
Application of Derivatives
Various Applications of Derivatives-
Rate of Change of Quantities:
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.
Increasing and Decreasing Function:
Consider y = f(x) be a differentiable function (whose derivative exists at all points in the domain) in an interval x = (a,b).
- If for any two points x1 and x2 in the interval x such a manner that x1 < x2, there holds an inequality f(x1) ≤ f(x2); then the function f(x) is known as increasing in this interval.
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- The functions are commonly known as strictly increasing or decreasing functions, given the inequalities are strict: f(x1) < f(x2) for strictly increasing and f(x1) > f(x2) for strictly decreasing.
Read More: Application of Derivatives