English

By Computing the Shortest Distance Determine Whether the Following Pairs of Lines Intersect Or Not: X − 1 2 = Y + 1 3 = Z a N D X + 1 5 = Y − 2 1 ; Z = 2

Advertisements
Advertisements

Question

By computing the shortest distance determine whether the following pairs of lines intersect or not: \[\frac{x - 1}{2} = \frac{y + 1}{3} = z \text{ and } \frac{x + 1}{5} = \frac{y - 2}{1}; z = 2\]

Sum
Advertisements

Solution

\[\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 0}{1} and \frac{x + 1}{5} = \frac{y - 2}{1} = \frac{z - 2}{0}\]

Since the first line passes through the point (1, -1, 0) and has direction ratios proportional to 2, 3, 1, its vector equation is

\[\overrightarrow{r} = \overrightarrow{a_1} + \lambda \overrightarrow{b_1} . . . (1) \]

\[\text{ Here } , \]

\[ \overrightarrow{a_1} = \hat{i}  - \hat{j}  + 0 \hat{k}  \]

\[ \overrightarrow{b_1} = 2 \hat{i}  + 3 \hat{j}  + \hat{k} \]

Also, the second line passes through the point ( -1, 2, 2) and has direction ratios proportional to 5, 1, 0. Its vector equation is

\[\overrightarrow{r} = \overrightarrow{a_2} + \mu \overrightarrow{b_2} . . . (2) \]

\[\text{ Here }, \]

\[ \overrightarrow{a_2} = - \hat{i} + 2 \hat{j} + 2 \hat{k} \]

\[ \overrightarrow{b_2} = 5 \hat{i} + \hat{j}  + 0 \hat{k} \]

Now,

\[\overrightarrow{a_2} - \overrightarrow{a_1} = - 2 \hat{i} + 3 \hat{j}  + 2 \hat{k}  \]

\[\text{ and } \overrightarrow{b_1} \times \overrightarrow{b_2} = \begin{vmatrix}\hat{i}  & \hat{j}  & \hat{k}  \\ 2 & 3 & 1 \\ 5 & 1 & 0\end{vmatrix}\]

\[ = - \hat{i}  + 5 \hat{j} - 13 \hat{k}  \]

\[\left( \overrightarrow{a_2} - \overrightarrow{a_1} \right) . \left( \overrightarrow{b_1} \times \overrightarrow{b_2} \right) = \left( - 2 \hat{i} + 3 \hat{j} + 2 \hat{k} \right) . \left( - \hat{i} + 5 \hat{j} - 13 \hat{k} \right)\]

\[ = 2 + 15 - 26\]

\[ = - 9\]

\[\text{ We observe } \]

\[\left( \overrightarrow{a_2} - \overrightarrow{a_1} \right) . \left( \overrightarrow{b_1} \times \overrightarrow{b_2} \right) \neq 0\]

\[\text{ Thus, the given lines do not intersect } .\]

shaalaa.com
  Is there an error in this question or solution?
Chapter 27: Straight Line in Space - Exercise 28.5 [Page 38]

APPEARS IN

R.D. Sharma Mathematics Volume 1 and 2 [English] Class 12
Chapter 27 Straight Line in Space
Exercise 28.5 | Q 3.3 | Page 38

RELATED QUESTIONS

The Cartesian equations of line are 3x -1 = 6y + 2 = 1 - z. Find the vector equation of line.


 

Find the value of p, so that the lines `l_1:(1-x)/3=(7y-14)/p=(z-3)/2 and l_2=(7-7x)/3p=(y-5)/1=(6-z)/5 ` are perpendicular to each other. Also find the equations of a line passing through a point (3, 2, – 4) and parallel to line l1.

 

Let `A(bara)` and `B(barb)` be any two points in the space and `R(barr)` be a point on the line segment AB dividing it internally in the ratio m : n, then prove that `bar r=(mbarb+nbara)/(m+n)`. Hence find the position vector of R which divides the line segment joining the points A(1, –2, 1) and B(1, 4, –2) internally in the ratio 2 : 1.


Find the vector equation of the lines which passes through the point with position vector `4hati - hatj +2hatk` and is in the direction of `-2hati + hatj + hatk`


Find the vector equation for the line which passes through the point (1, 2, 3) and parallel to the vector \[\hat{i} - 2 \hat{j} + 3 \hat{k} .\]  Reduce the corresponding equation in cartesian from.


The cartesian equations of a line are x = ay + bz = cy + d. Find its direction ratios and reduce it to vector form. 


Show that the three lines with direction cosines \[\frac{12}{13}, \frac{- 3}{13}, \frac{- 4}{13}; \frac{4}{13}, \frac{12}{13}, \frac{3}{13}; \frac{3}{13}, \frac{- 4}{13}, \frac{12}{13}\] are mutually perpendicular. 


Find the angle between the following pair of line:

\[\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{- 3} \text { and } \frac{x + 3}{- 1} = \frac{y - 5}{8} = \frac{z - 1}{4}\]


Find the angle between the following pair of line:

\[\frac{- x + 2}{- 2} = \frac{y - 1}{7} = \frac{z + 3}{- 3} \text{  and  } \frac{x + 2}{- 1} = \frac{2y - 8}{4} = \frac{z - 5}{4}\]


Find the equation of the line passing through the point  \[\hat{i}  + \hat{j}  - 3 \hat{k} \] and perpendicular to the lines  \[\overrightarrow{r} = \hat{i}  + \lambda\left( 2 \hat{i} + \hat{j}  - 3 \hat{k}  \right) \text { and }  \overrightarrow{r} = \left( 2 \hat{i}  + \hat{j}  - \hat{ k}  \right) + \mu\left( \hat{i}  + \hat{j}  + \hat{k}  \right) .\]

  

 

 

 


Show that the lines \[\frac{x - 1}{3} = \frac{y + 1}{2} = \frac{z - 1}{5} \text{           and                } \frac{x + 2}{4} = \frac{y - 1}{3} = \frac{z + 1}{- 2}\]  do not intersect. 


Show that the lines \[\frac{x + 1}{3} = \frac{y + 3}{5} = \frac{z + 5}{7} \text{           and                  } \frac{x - 2}{1} = \frac{y - 4}{3} = \frac{z - 6}{5}\]   intersect. Find their point of intersection.


Prove that the line \[\vec{r} = \left( \hat{i }+ \hat{j }- \hat{k} \right) + \lambda\left( 3 \hat{i} - \hat{j} \right) \text{ and } \vec{r} = \left( 4 \hat{i} - \hat{k} \right) + \mu\left( 2 \hat{i} + 3 \hat{k} \right)\] intersect and find their point of intersection.


Find the foot of the perpendicular from (1, 2, −3) to the line \[\frac{x + 1}{2} = \frac{y - 3}{- 2} = \frac{z}{- 1} .\]


Find the coordinates of the foot of perpendicular drawn from the point A(1, 8, 4) to the line joining the points B(0, −1, 3) and C(2, −3, −1).      


Find the shortest distance between the following pairs of lines whose vector equations are: \[\overrightarrow{r} = \left( 1 - t \right) \hat{i} + \left( t - 2 \right) \hat{j} + \left( 3 - t \right) \hat{k}  \text{ and }  \overrightarrow{r} = \left( s + 1 \right) \hat{i}  + \left( 2s - 1 \right) \hat{j}  - \left( 2s + 1 \right) \hat{k} \]


By computing the shortest distance determine whether the following pairs of lines intersect or not  : \[\overrightarrow{r} = \left( \hat{i} - \hat{j} \right) + \lambda\left( 2 \hat{i} + \hat{k}  \right) \text{ and }  \overrightarrow{r} = \left( 2 \hat{i} - \hat{j}  \right) + \mu\left( \hat{i} + \hat{j} - \hat{k} \right)\]


Find the shortest distance between the following pairs of parallel lines whose equations are: \[\overrightarrow{r} = \left( \hat{i} + \hat{j} \right) + \lambda\left( 2 \hat{i} - \hat{j} + \hat{k} \right) \text{ and } \overrightarrow{r} = \left( 2 \hat{i} + \hat{j} - \hat{k} \right) + \mu\left( 4 \hat{i} - 2 \hat{j} + 2 \hat{k} \right)\]


Find the equations of the lines joining the following pairs of vertices and then find the shortest distance between the lines
(i) (0, 0, 0) and (1, 0, 2) 


Find the equations of the lines joining the following pairs of vertices and then find the shortest distance between the lines

 (1, 3, 0) and (0, 3, 0)


Find the shortest distance between the lines \[\overrightarrow{r} = \hat{i} + 2 \hat{j} + 3 \hat{k} + \lambda\left( \hat{i} - 3 \hat{j} + 2 \hat{k} \right) \text{ and }  \overrightarrow{r} = 4 \hat{i} + 5 \hat{j}  + 6 \hat{k} + \mu\left( 2 \hat{i} + 3 \hat{j} + \hat{k} \right)\]


Find the shortest distance between the lines \[\overrightarrow{r} = 6 \hat{i} + 2 \hat{j} + 2 \hat{k} + \lambda\left( \hat{i} - 2 \hat{j} + 2 \hat{k} \right) \text{ and }  \overrightarrow{r} = - 4 \hat{i}  - \hat{k}  + \mu\left( 3 \hat{i} - 2 \hat{j} - 2 \hat{k}  \right)\]


Write the cartesian and vector equations of X-axis.

 

Write the cartesian and vector equations of Y-axis.

 

Write the vector equation of a line passing through a point having position vector  \[\overrightarrow{\alpha}\] and parallel to vector \[\overrightarrow{\beta}\] .


Write the coordinate axis to which the line \[\frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 1}{0}\]  is  perpendicular.


Write the direction cosines of the line whose cartesian equations are 2x = 3y = −z.

 

Find the angle between the lines 

\[\vec{r} = \left( 2 \hat{i}  - 5 \hat{j}  + \hat{k}  \right) + \lambda\left( 3 \hat{i}  + 2 \hat{j}  + 6 \hat{k}  \right)\] and \[\vec{r} = 7 \hat{i} - 6 \hat{k}  + \mu\left( \hat{i}  + 2 \hat{j}  + 2 \hat{k}  \right)\] 


The projections of a line segment on XY and Z axes are 12, 4 and 3 respectively. The length and direction cosines of the line segment are


If y – 2x – k = 0 touches the conic 3x2 – 5y2 = 15, find the value of k. 


Auxillary equation of 2x2 + 3xy − 9y2 = 0 is ______ 


Find the position vector of a point A in space such that `vec"OA"` is inclined at 60º to OX and at 45° to OY and `|vec"OA"|` = 10 units.


The lines `(x - 1)/2 = (y + 1)/2 = (z - 1)/4` and `(x - 3)/1 = (y - k)/2 = z/1` intersect each other at point


A line passes through \[A(x_1,y_1,z_1)\] and has direction ratios \[a,b,c\]. Which is its Cartesian equation?


What is the parametric form of the line through \[(5,2,-4)\] with direction ratios \[3,2,-8\]?


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×