The shortest distance between lines bar{r}=(h | vedclass

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

Question

(D) The given lines are $\bar{r} = \bar{a}_1 + \lambda \bar{b}_1$ and $\bar{r} = \bar{a}_2 + \mu \bar{b}_2$. Here,$\bar{a}_1 = \hat{i} + 2\hat{j} - \h

Vedclass · Accepted Answer

$\frac{2 \sqrt{2}}{\sqrt{19}}$ units

Vedclass · Answer

(D) The given lines are $\bar{r} = \bar{a}_1 + \lambda \bar{b}_1$ and $\bar{r} = \bar{a}_2 + \mu \bar{b}_2$. Here,$\bar{a}_1 = \hat{i} + 2\hat{j} - \hat{k}$,$\bar{b}_1 = 2\hat{i} + \hat{j} - 3\hat{k}$,$\bar{a}_2 = 2\hat{i} - \hat{j} + 2\hat{k}$,and $\bar{b}_2 = \hat{i} - \hat{j} + \hat{k}$. First,calculate $\bar{a}_2 - \bar{a}_1 = (2-1)\hat{i} + (-1-2)\hat{j} + (2-(-1))\hat{k} = \hat{i} - 3\hat{j} + 3\hat{k}$. Next,calculate the cross product $\bar{b}_1 	imes \bar{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 1 & -3 \ 1 & -1 & 1 \end{vmatrix} = \hat{i}(1-3) - \hat{j}(2+3) + \hat{k}(-2-1) = -2\hat{i} - 5\hat{j} - 3\hat{k}$. The magnitude is $|\bar{b}_1 	imes \bar{b}_2| = \sqrt{(-2)^2 + (-5)^2 + (-3)^2} = \sqrt{4 + 25 + 9} = \sqrt{38}$. The shortest distance is $d = \left| \frac{(\bar{b}_1 	imes \bar{b}_2) \cdot (\bar{a}_2 - \bar{a}_1)}{|\bar{b}_1 	imes \bar{b}_2|} ight| = \left| \frac{(-2\hat{i} - 5\hat{j} - 3\hat{k}) \cdot (\hat{i} - 3\hat{j} + 3\hat{k})}{\sqrt{38}} ight| = \left| \frac{-2 + 15 - 9}{\sqrt{38}} ight| = \frac{4}{\sqrt{38}} = \frac{4}{\sqrt{2} 	imes \sqrt{19}} = \frac{2 	imes 2}{\sqrt{2} 	imes \sqrt{19}} = \frac{2 \sqrt{2}}{\sqrt{19}}$ units.

The shortest distance between lines $\bar{r}=(\hat{i}+2 \hat{j}-\hat{k})+\lambda(2 \hat{i}+\hat{j}-3 \hat{k})$ and $\bar{r}=(2 \hat{i}-\hat{j}+2 \hat{k})+\mu(\hat{i}-\hat{j}+\hat{k})$ is

Similar Questions

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

If the line passes through the points $P(6, -1, 2)$,$Q(8, -7, 2\lambda)$,and $R(5, 2, 4)$,then the value of $\lambda$ is $.......$

Given the lines $\vec{r} = (3+t)\hat{i} + (1-t)\hat{j} + (-2-2t)\hat{k}$,$t \in R$ and $x = 4+k, y = -k, z = -4-2k$,$k \in R$. What is the relationship between these two lines?

If the square of the shortest distance between the lines $\frac{x-2}{1}=\frac{y-1}{2}=\frac{z+3}{-3}$ and $\frac{x+1}{2}=\frac{y+3}{4}=\frac{z+5}{-5}$ is $\frac{m}{n}$,where $m, n$ are coprime numbers,then $m+n$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module