If the square of the shortest distance betwee | vedclass

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

Question

(B) The lines are given by $\vec{r} = \vec{a_1} + \lambda \vec{p}$ and $\vec{r} = \vec{a_2} + \mu \vec{q}$,where $\vec{a_1} = (2, 1, -3)$,$\vec{p} = (

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(B) The lines are given by $\vec{r} = \vec{a_1} + \lambda \vec{p}$ and $\vec{r} = \vec{a_2} + \mu \vec{q}$,where $\vec{a_1} = (2, 1, -3)$,$\vec{p} = (1, 2, -3)$,$\vec{a_2} = (-1, -3, -5)$,and $\vec{q} = (2, 4, -5)$.The shortest distance $d$ between two skew lines is given by $d = \frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{p} 	imes \vec{q})|}{|\vec{p} 	imes \vec{q}|}$.First,calculate $\vec{p} 	imes \vec{q} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 2 & -3 \ 2 & 4 & -5 \end{vmatrix} = \hat{i}(-10 + 12) - \hat{j}(-5 + 6) + \hat{k}(4 - 4) = 2\hat{i} - \hat{j} + 0\hat{k}$.The magnitude is $|\vec{p} 	imes \vec{q}| = \sqrt{2^2 + (-1)^2 + 0^2} = \sqrt{5}$.Next,$\vec{a_2} - \vec{a_1} = (-1-2, -3-1, -5-(-3)) = (-3, -4, -2)$.The dot product $(\vec{a_2} - \vec{a_1}) \cdot (\vec{p} 	imes \vec{q}) = (-3, -4, -2) \cdot (2, -1, 0) = -6 + 4 + 0 = -2$.Thus,$d = \frac{|-2|}{\sqrt{5}} = \frac{2}{\sqrt{5}}$.The square of the shortest distance is $d^2 = \frac{4}{5}$.Given $d^2 = \frac{m}{n} = \frac{4}{5}$,where $m=4$ and $n=5$ are coprime.Therefore,$m+n = 4+5 = 9$.

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:

Similar Questions

The Cartesian equation of a line is $\frac{x-5}{3}=\frac{y+4}{7}=\frac{6-z}{2}$. Find the vector equation of the line.

What is the equation of the line passing through the point $(a, b, c)$ and parallel to the $z$-axis?

If the lines $\frac{x - 1}{k} = \frac{y - 2}{2} = \frac{z - 3}{3}$ and $\frac{x - 2}{3} = \frac{y - 3}{k} = \frac{z - 1}{2}$ intersect,find the value of $k$.

The shortest distance between the lines $\frac{x - 3}{2} = \frac{y + 15}{-7} = \frac{z - 9}{5}$ and $\frac{x + 1}{2} = \frac{y - 1}{1} = \frac{z - 9}{-3}$ is

$A$ line having direction ratios $1, -4, 2$ intersects the lines $\frac{x-7}{3} = \frac{y-1}{-1} = \frac{z+2}{1}$ and $\frac{x}{2} = \frac{y-7}{3} = \frac{z}{1}$ at the points $A$ and $B$ respectively. Then,the coordinates of points $A$ and $B$ are:

Vedclass Test Series

Exam Paper Generator

Online Exam Module