The shortest distance between the lines frac{ | vedclass

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

Question

(A) The shortest distance $d$ between two lines $\vec{r} = \vec{a_1} + \lambda \vec{b_1}$ and $\vec{r} = \vec{a_2} + \mu \vec{b_2}$ is given by $d = \

Vedclass · Accepted Answer

$\frac{2}{\sqrt{69}}$ units

Vedclass · Answer

(A) The shortest distance $d$ between two lines $\vec{r} = \vec{a_1} + \lambda \vec{b_1}$ and $\vec{r} = \vec{a_2} + \mu \vec{b_2}$ is given by $d = \frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} 	imes \vec{b_2})|}{ |\vec{b_1} 	imes \vec{b_2}| }$. For the given lines: Line $1$: $\vec{a_1} = (-1, 2, -1)$,$\vec{b_1} = (3, 2, 2)$. Line $2$: $\vec{a_2} = (2, 2, -3)$,$\vec{b_2} = (1, 2, 3)$. $\vec{a_2} - \vec{a_1} = (2 - (-1), 2 - 2, -3 - (-1)) = (3, 0, -2)$. $\vec{b_1} 	imes \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & 2 & 2 \ 1 & 2 & 3 \end{vmatrix} = \hat{i}(6 - 4) - \hat{j}(9 - 2) + \hat{k}(6 - 2) = 2\hat{i} - 7\hat{j} + 4\hat{k}$. $|\vec{b_1} 	imes \vec{b_2}| = \sqrt{2^2 + (-7)^2 + 4^2} = \sqrt{4 + 49 + 16} = \sqrt{69}$. $(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} 	imes \vec{b_2}) = (3)(2) + (0)(-7) + (-2)(4) = 6 + 0 - 8 = -2$. $d = \frac{|-2|}{\sqrt{69}} = \frac{2}{\sqrt{69}}$ units.

The shortest distance between the lines $\frac{x+1}{3}=\frac{y-2}{2}=\frac{z+1}{2}$ and $\frac{x-2}{1}=\frac{y-2}{2}=\frac{z+3}{3}$ is

Similar Questions

If the image of the point $P(1, 0, 3)$ in the line joining the points $A(4, 7, 1)$ and $B(3, 5, 3)$ is $Q(\alpha, \beta, \gamma)$,then $\alpha + \beta + \gamma$ is equal to

Find the Cartesian equation of the line passing through the point $\hat{i} + 2\hat{j} + 2\hat{k}$ and parallel to the line joining the points $2\hat{i} - \hat{j} + \hat{k}$ and $-\hat{i} + 4\hat{j} + \hat{k}$.

Find the equation of the line in vector and in Cartesian form that passes through the point with position vector $2\hat{i}-\hat{j}+4\hat{k}$ and is in the direction $\hat{i}+2\hat{j}-\hat{k}$.

The square of the distance of the point $\left(\frac{15}{7}, \frac{32}{7}, 7\right)$ from the line $\frac{x+1}{3}=\frac{y+3}{5}=\frac{z+5}{7}$ in the direction of the vector $\hat{i}+4 \hat{j}+7 \hat{k}$ is :

The angle between the lines $\vec{r}=(2 \hat{i}+\hat{j}-3 \hat{k})+\lambda(\hat{i}-\hat{j}+\hat{k})$ and $\frac{x-1}{1}=\frac{y+2}{3}=\frac{z-3}{2}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module