Let L_1: frac{x-1}{2}=frac{y-2}{3}=frac{z-3}{ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Let $P(2\lambda+1, 3\lambda+2, 4\lambda+3)$ be a point on $L_1$ and $Q(3\mu+2, 4\mu+4, 5\mu+5)$ be a point on $L_2$.The direction ratios of $PQ$ a

Vedclass · Accepted Answer

$\left(\frac{14}{3},-3, \frac{22}{3}\right)$

Vedclass · Answer

(D) Let $P(2\lambda+1, 3\lambda+2, 4\lambda+3)$ be a point on $L_1$ and $Q(3\mu+2, 4\mu+4, 5\mu+5)$ be a point on $L_2$.The direction ratios of $PQ$ are $(3\mu-2\lambda+1, 4\mu-3\lambda+2, 5\mu-4\lambda+2)$.Since $PQ$ is the shortest distance line,$PQ \perp L_1$ and $PQ \perp L_2$.For $PQ \perp L_1$: $2(3\mu-2\lambda+1) + 3(4\mu-3\lambda+2) + 4(5\mu-4\lambda+2) = 0 \Rightarrow 38\mu - 29\lambda + 16 = 0$.For $PQ \perp L_2$: $3(3\mu-2\lambda+1) + 4(4\mu-3\lambda+2) + 5(5\mu-4\lambda+2) = 0 \Rightarrow 50\mu - 38\lambda + 21 = 0$.Solving these equations,we get $\lambda = \frac{1}{3}$ and $\mu = -\frac{1}{6}$.Substituting these values,$P = \left(\frac{5}{3}, 3, \frac{13}{3}\right)$ and $Q = \left(\frac{3}{2}, \frac{10}{3}, \frac{25}{6}\right)$.The equation of the line $PQ$ passing through $P$ with direction ratios proportional to $\vec{PQ} = Q-P = \left(-\frac{1}{6}, \frac{1}{3}, -\frac{1}{6}\right)$ (or $(1, -2, 1)$) is $\frac{x-5/3}{1} = \frac{y-3}{-2} = \frac{z-13/3}{1}$.Checking the options,the point $\left(\frac{14}{3}, -3, \frac{22}{3}\right)$ satisfies this equation: $\frac{14/3 - 5/3}{1} = 3$,$\frac{-3-3}{-2} = 3$,$\frac{22/3 - 13/3}{1} = 3$.

Let $L_1: \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $L_2: \frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5}$ be two lines. Then which of the following points lies on the line of the shortest distance between $L_1$ and $L_2$?

Similar Questions

If the direction cosines of two lines are given by $l+m+n=0$ and $mn-2lm-2nl=0$,then the acute angle between those lines is

Direction ratios of the line represented by the equations $x = ay + b$ and $z = cy + d$ are:

Let $a, b \in R$. If the mirror image of the point $P(a, 6, 9)$ with respect to the line $\frac{x-3}{7} = \frac{y-2}{5} = \frac{z-1}{-9}$ is $(20, b, -a-9)$,then $|a+b|$ is equal to

The equation of the line passing through the point $Q(0,1,2)$ and perpendicular to the line $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{-2}$ is

The distance of the point $B(1, 2, 3)$ from the line passing through $A(4, 2, 2)$ and parallel to the vector $\vec{c} = 2i + 3j + 6k$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module