Consider the lines L_1 and L_2 given byL_1: f | vedclass

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

Question

(A) Let the point $P$ on $L_1$ be $(2\lambda+1, \lambda+3, 2\lambda+2)$ and the point $Q$ on $L_2$ be $(\mu+2, 2\mu+2, 3\mu+3)$.The direction ratios o

Vedclass · Accepted Answer

$2\sqrt{6}$

Vedclass · Answer

(A) Let the point $P$ on $L_1$ be $(2\lambda+1, \lambda+3, 2\lambda+2)$ and the point $Q$ on $L_2$ be $(\mu+2, 2\mu+2, 3\mu+3)$.The direction ratios of the line $PQ$ are given by $((\mu+2)-(2\lambda+1), (2\mu+2)-(\lambda+3), (3\mu+3)-(2\lambda+2)) = (\mu-2\lambda+1, 2\mu-\lambda-1, 3\mu-2\lambda+1)$.Since the direction ratios of $L_3$ are $1, -1, -2$,we have:$\frac{\mu-2\lambda+1}{1} = \frac{2\mu-\lambda-1}{-1} = \frac{3\mu-2\lambda+1}{-2}$.From the first two parts: $-\mu+2\lambda-1 = 2\mu-\lambda-1 \Rightarrow 3\lambda = 3\mu \Rightarrow \lambda = \mu$.Substituting $\lambda = \mu$ into the equality $\frac{\mu-2\lambda+1}{1} = \frac{3\mu-2\lambda+1}{-2}$:$\frac{-\lambda+1}{1} = \frac{\lambda+1}{-2} \Rightarrow 2\lambda - 2 = \lambda + 1 \Rightarrow \lambda = 3$.Thus,$\lambda = 3$ and $\mu = 3$. The points are $P(7, 6, 8)$ and $Q(5, 8, 12)$.The length $PQ = \sqrt{(5-7)^2 + (8-6)^2 + (12-8)^2} = \sqrt{(-2)^2 + 2^2 + 4^2} = \sqrt{4 + 4 + 16} = \sqrt{24} = 2\sqrt{6}$.

Similar Questions

Let $P$ be the foot of the perpendicular from the point $Q(10,-3,-1)$ on the line $\frac{x-3}{7}=\frac{y-2}{-1}=\frac{z+1}{-2}$. Then the area of the right-angled triangle $PQR$,where $R$ is the point $(3,-2,1)$,is

The shortest distance between the lines $\overline{r}=(3 \bar{i}-5 \bar{j}+2 \bar{k})+t(4 \bar{i}+3 \bar{j}-\bar{k})$ and $\overline{r}=(\bar{i}+2 \bar{j}-4 \bar{k})+s(6 \bar{i}+3 \bar{j}-2 \bar{k})$ is

Find the angle between the lines $\vec{r}=3 \hat{i}-2 \hat{j}+6 \hat{k}+\lambda(2 \hat{i}+\hat{j}+2 \hat{k})$ and $\vec{r}=(2 \hat{j}-5 \hat{k})+\mu(6 \hat{i}+3 \hat{j}+2 \hat{k})$.

If lines $\frac{2x-4}{\lambda}=\frac{y-1}{2}=\frac{z-3}{1}$ and $\frac{x-1}{1}=\frac{3y-1}{\lambda}=\frac{z-2}{1}$ are perpendicular to each other,then $\lambda = \ldots$.

Let $(\alpha, \beta, \gamma)$ be the foot of the perpendicular from the point $(1, 2, 3)$ on the line $\frac{x+3}{5} = \frac{y-1}{2} = \frac{z+4}{3}$. Then $19(\alpha + \beta + \gamma)$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module