The shortest distance between the skew lines | vedclass

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

Question

(D) The formula for the shortest distance between two skew lines $\vec{r} = \vec{a_1} + t\vec{b_1}$ and $\vec{r} = \vec{a_2} + s\vec{b_2}$ is given by

Vedclass · Accepted Answer

$\sqrt{3}$

Vedclass · Answer

(D) The formula for the shortest distance between two skew lines $\vec{r} = \vec{a_1} + t\vec{b_1}$ and $\vec{r} = \vec{a_2} + s\vec{b_2}$ is given by $d = \frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} 	imes \vec{b_2})|}{ |\vec{b_1} 	imes \vec{b_2}| }$.Given $\vec{a_1} = \hat{i} + 2\hat{j} + 3\hat{k}$,$\vec{b_1} = \hat{i} + 3\hat{j} + 2\hat{k}$,$\vec{a_2} = 4\hat{i} + 5\hat{j} + 6\hat{k}$,and $\vec{b_2} = 2\hat{i} + 3\hat{j} + \hat{k}$.First,calculate $\vec{a_2} - \vec{a_1} = (4-1)\hat{i} + (5-2)\hat{j} + (6-3)\hat{k} = 3\hat{i} + 3\hat{j} + 3\hat{k}$.Next,calculate the cross product $\vec{b_1} 	imes \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 3 & 2 \ 2 & 3 & 1 \end{vmatrix} = \hat{i}(3-6) - \hat{j}(1-4) + \hat{k}(3-6) = -3\hat{i} + 3\hat{j} - 3\hat{k}$.The magnitude $|\vec{b_1} 	imes \vec{b_2}| = \sqrt{(-3)^2 + 3^2 + (-3)^2} = \sqrt{9 + 9 + 9} = \sqrt{27} = 3\sqrt{3}$.Now,the dot product $(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} 	imes \vec{b_2}) = (3\hat{i} + 3\hat{j} + 3\hat{k}) \cdot (-3\hat{i} + 3\hat{j} - 3\hat{k}) = -9 + 9 - 9 = -9$.The shortest distance $d = \frac{|-9|}{3\sqrt{3}} = \frac{9}{3\sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3}$.

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

Similar Questions

If the line $\frac{2-x}{3}=\frac{3y-2}{4\lambda+1}=4-z$ makes a right angle with the line $\frac{x+3}{3\mu}=\frac{1-2y}{6}=\frac{5-z}{7}$,then $4\lambda+9\mu$ is equal to :

The perimeter of a square whose two sides lie along the lines $\frac{x-1}{2}=\frac{y+2}{3}=\frac{z-3}{4}$ and $\frac{x}{2}=\frac{y-1}{3}=\frac{z+1}{4}$ is

The Cartesian equation of a line is $2x - 3 = 3y + 1 = 5 - 6z$. The vector equation of the line passing through the point $(7, -5, 0)$ and parallel to the given line is

If the $x$-coordinate of a point $P$ on the line joining the points $Q(2, 2, 1)$ and $R(5, 2, -2)$ is $4$,then the $y$-coordinate of $P$ is:

If the image of the point $P(1, 2, a)$ in the line $\frac{x-6}{3}=\frac{y-7}{2}=\frac{7-z}{2}$ is $Q(5, b, c)$,then $a^{2}+b^{2}+c^{2}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module