The shortest distance between the skew lines | vedclass

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

Question

(C) The shortest distance $d$ between two skew lines $\vec{r}=\vec{a}_1+\lambda \vec{b}_1$ and $\vec{r}=\vec{a}_2+\mu \vec{b}_2$ is given by the formu

Vedclass · Accepted Answer

$\sqrt{35}$

Vedclass · Answer

(C) The shortest distance $d$ between two skew lines $\vec{r}=\vec{a}_1+\lambda \vec{b}_1$ and $\vec{r}=\vec{a}_2+\mu \vec{b}_2$ is given by the formula: $d = \left| \frac{(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 	imes \vec{b}_2)}{|\vec{b}_1 	imes \vec{b}_2|} ight|$.Given lines are $\vec{r}=(3 \hat{i}+4 \hat{j}-2 \hat{k})+\lambda(-\hat{i}+2 \hat{j}+\hat{k})$ and $\vec{r}=(\hat{i}-7 \hat{j}-2 \hat{k})+\mu(\hat{i}+3 \hat{j}+2 \hat{k})$.Here, $\vec{a}_1 = 3 \hat{i}+4 \hat{j}-2 \hat{k}$, $\vec{b}_1 = -\hat{i}+2 \hat{j}+\hat{k}$, $\vec{a}_2 = \hat{i}-7 \hat{j}-2 \hat{k}$, and $\vec{b}_2 = \hat{i}+3 \hat{j}+2 \hat{k}$.First, calculate $\vec{a}_2 - \vec{a}_1 = (1-3)\hat{i} + (-7-4)\hat{j} + (-2-(-2))\hat{k} = -2\hat{i} - 11\hat{j}$.Next, calculate the cross product $\vec{b}_1 	imes \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -1 & 2 & 1 \ 1 & 3 & 2 \end{vmatrix} = \hat{i}(4-3) - \hat{j}(-2-1) + \hat{k}(-3-2) = \hat{i} + 3\hat{j} - 5\hat{k}$.The magnitude $|\vec{b}_1 	imes \vec{b}_2| = \sqrt{1^2 + 3^2 + (-5)^2} = \sqrt{1+9+25} = \sqrt{35}$.Now, the dot product $(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 	imes \vec{b}_2) = (-2\hat{i} - 11\hat{j} + 0\hat{k}) \cdot (\hat{i} + 3\hat{j} - 5\hat{k}) = (-2)(1) + (-11)(3) + (0)(-5) = -2 - 33 = -35$.Finally, $d = \left| \frac{-35}{\sqrt{35}} ight| = \sqrt{35}$.

The shortest distance between the skew lines $\vec{r}=(3 \hat{i}+4 \hat{j}-2 \hat{k})+\lambda(-\hat{i}+2 \hat{j}+\hat{k})$ and $\vec{r}=(\hat{i}-7 \hat{j}-2 \hat{k})+\mu(\hat{i}+3 \hat{j}+2 \hat{k})$ is

Similar Questions

The length of the perpendicular drawn from the point $P(5, 4, -1)$ to the line $\frac{x - 1}{2} = \frac{y}{9} = \frac{z}{5}$ is:

The equation of the line passing through $(3, -1, 2)$ and perpendicular to the lines $\vec{r} = (\hat{i} + \hat{j} - \hat{k}) + \lambda(2 \hat{i} - 2 \hat{j} + \hat{k})$ and $\vec{r} = (2 \hat{i} + \hat{j} - 3 \hat{k}) + \mu(\hat{i} - 2 \hat{j} + 2 \hat{k})$ is

Let a triangle $PQR$ be such that $P$ and $Q$ lie on the line $\frac{x+3}{8} = \frac{y-4}{2} = \frac{z+1}{2}$ and are at a distance of $6$ units from $R(1, 2, 3)$. If $(\alpha, \beta, \gamma)$ is the centroid of $\triangle PQR$, then $\alpha + \beta + \gamma$ is equal to :

The shortest distance between the lines $r = (3i - 2j - 2k) + t(i)$ and $r = (i - j + 2k) + s(j)$ ($t$ and $s$ being parameters) is

The Cartesian equation of a line is $\frac{x+2}{3}=\frac{y-4}{2}=\frac{z-5}{5}$,then the vector equation of the line is

Vedclass Test Series

Exam Paper Generator

Online Exam Module