Let L_1 and L_2 denote the lines overrightarr | vedclass

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

Question

(A) The lines are $L_1: \overrightarrow{r} = (1, 0, 0) + \lambda(-1, 2, 2)$ and $L_2: \overrightarrow{r} = \mu(2, -1, 2)$.Let $A$ be a point on $L_1$

Vedclass · Accepted Answer

$1, 2, 4$

Vedclass · Answer

(A) The lines are $L_1: \overrightarrow{r} = (1, 0, 0) + \lambda(-1, 2, 2)$ and $L_2: \overrightarrow{r} = \mu(2, -1, 2)$.Let $A$ be a point on $L_1$ and $B$ be a point on $L_2$. $A = (1-\lambda, 2\lambda, 2\lambda)$ and $B = (2\mu, -\mu, 2\mu)$.The vector $\overrightarrow{AB} = (2\mu + \lambda - 1, -\mu - 2\lambda, 2\mu - 2\lambda)$.The direction of the common perpendicular is $\vec{v} = (-1, 2, 2) 	imes (2, -1, 2) = (6, 6, -6)$,which is parallel to $(2, 2, -1)$.Since $AB$ is the shortest distance line,$\overrightarrow{AB}$ must be parallel to $(2, 2, -1)$.Thus,$\frac{2\mu + \lambda - 1}{2} = \frac{-\mu - 2\lambda}{2} = \frac{2\mu - 2\lambda}{-1} = k$.Solving this system gives $\lambda = 1/9$ and $\mu = 2/9$.Then $A = (8/9, 2/9, 2/9)$ and $B = (4/9, -2/9, 4/9)$.The line $L_3$ passes through $A$ and $B$ with direction $(2, 2, -1)$.Equation of $L_3$: $\overrightarrow{r} = A + t(2, 2, -1) = (8/9, 2/9, 2/9) + t(2, 2, -1)$.Option $(1)$ passes through $(2/3, 0, 1/3)$,which is the midpoint of $AB$. Since the direction is $(2, 2, -1)$,it represents $L_3$.Option $(2)$ passes through $B$,so it represents $L_3$.Option $(4)$ passes through $(8/9, 2/9, 2/9)$,which is point $A$,so it represents $L_3$.Thus,options $1, 2, 4$ are correct.

Similar Questions

The direction cosines of a line which is perpendicular to the lines $\frac{x - 7}{2} = \frac{y + 17}{-3} = \frac{z - 6}{1}$ and $\frac{x + 5}{1} = \frac{y + 3}{2} = \frac{z - 6}{-2}$ are...

Find the point of intersection of the lines $\vec{r} = (\hat{i} + \hat{j} - \hat{k}) + \lambda(3\hat{i} - \hat{j})$ and $\vec{r} = (4\hat{i} - \hat{k}) + \mu(2\hat{i} + 3\hat{k})$.

The equation of a line passing through the point $(2,1,3)$ and perpendicular to the lines $\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{3}$ and $\frac{x}{-3}=\frac{y}{2}=\frac{z}{5}$ is

The angle between two lines $\frac{x+1}{2}=\frac{y+3}{2}=\frac{z-4}{-1}$ and $\frac{x-4}{1}=\frac{y+4}{2}=\frac{z+1}{2}$ is

The lines $x = ay + b, z = cy + d$ and $x = a'y + b', z = c'y + d'$ are perpendicular to each other,if

Vedclass Test Series

Exam Paper Generator

Online Exam Module