English

Find the Shortest Distance Between the Lines → R = ^ I + 2 ^ J + 3 ^ K + λ ( ^ I − 3 ^ J + 2 ^ K ) and → R = 4 ^ I + 5 ^ J + 6 ^ K + μ ( 2 ^ I + 3 ^ J + ^ K ) - Mathematics

Advertisements
Advertisements

Question

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

Sum
Advertisements

Solution

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

Comparing the given equations with the equations 

\[\overrightarrow{r} = \overrightarrow{a_1} + \lambda \overrightarrow{b_1} \text{ and }  \overrightarrow{r} = \overrightarrow{a_2} + \mu \overrightarrow{b_2}\]

we get ,

\[\overrightarrow{a_1} = \hat{i}  + 2 \hat{j}  + 3 \hat{k} \]
\[ \overrightarrow{a_2} = 4 \hat{i} + 5 \hat{j} + 6 \hat{k} \]
\[ \overrightarrow{b_1} = \hat{i} - 3 \hat{j} + 2 \hat{k} \]
\[ \vec{b_2} = 2 \hat{i} + 3 \hat{j} + \hat{k} \]
\[ \therefore \overrightarrow{a_2} - \overrightarrow{a_1} = 3 \hat{i} + 3 \hat{j}  + 3 \hat{k}  \]
\[\text{ and  } \overrightarrow{b_1} \times \overrightarrow{b_2} = \begin{vmatrix}\hat{i}  & \hat{j}  & \hat{k}  \\ 1 & - 3 & 2 \\ 2 & 3 & 1\end{vmatrix}\]
\[ = - 9 \hat{i}  + 3 \hat{j}  + 9 \hat{k}  \]
\[ \Rightarrow \left| \overrightarrow{b_1} \times \overrightarrow{b_2} \right| = \sqrt{\left( - 9 \right)^2 + 3^2 + 9^2}\]
\[ = \sqrt{81 + 9 + 81}\]
\[ = \sqrt{171}\]
\[\text{ and }  \left( \overrightarrow{a_2} - \overrightarrow{a_1} \right) . \left( \overrightarrow{b_1} \times \overrightarrow{b_2} \right) = \left( 3 \hat{i} + 3 \hat{j} + 3 \hat{k} \right) . \left( - 9 \hat{i} + 3 \hat{j} + 9 \hat{k}  \right)\]
\[ = - 27 + 9 + 27\]
\[ = 9\]

The shortest distance between the lines 

\[\overrightarrow{r} = \overrightarrow{a_1} + \lambda \overrightarrow{b_1} \text{ and } \overrightarrow{r} = \overrightarrow{a_2} + \mu \overrightarrow{b_2}\]

\[d = \left| \frac{\left( \overrightarrow{a_2} - \overrightarrow{a_1} \right) . \left( \overrightarrow{b_1} \times \overrightarrow{b_2} \right)}{\left| \overrightarrow{b_1} \times \overrightarrow{b_2} \right|} \right|\]
\[ \Rightarrow d = \left| \frac{9}{\sqrt{171}} \right|\]
\[ = \frac{3}{\sqrt{19}}\]

shaalaa.com
  Is there an error in this question or solution?
Chapter 28: Straight Line in Space - Exercise 28.5 [Page 38]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 28 Straight Line in Space
Exercise 28.5 | Q 7.3 | 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.

 

Find the vector and cartesian equations of the line passing through the point (2, 1, 3) and perpendicular to the lines

`(x-1)/1=(y-2)/2=(z-3)/3 and x/(-3)=y/2=z/5`


Show that the line through the points (4, 7, 8) (2, 3, 4) is parallel to the line through the points (−1, −2, 1), (1, 2, 5).


Find the equation of the line which passes through the point (1, 2, 3) and is parallel to the vector `3hati+2hatj-2hatk`.


Find the Cartesian equation of the line which passes through the point (−2, 4, −5) and parallel to the line given by `(x+3)/3 = (y-4)/5 = (z+8)/6`.


Find the vector and the Cartesian equations of the line that passes through the points (3, −2, −5), (3, −2, 6).

 


Find the vector and Cartesian equations of a line passing through (1, 2, –4) and perpendicular to the two lines `(x - 8)/3 = (y + 19)/(-16) = (z - 10)/7` and `(x - 15)/3 = (y - 29)/8 = (z - 5)/(-5)`


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`


The cartesian equation of a line are 3x + 1 = 6y − 2 = 1 − z. Find the fixed point through which it passes, its direction ratios and also its vector equation.


Show that the lines \[\frac{x - 5}{7} = \frac{y + 2}{- 5} = \frac{z}{1} \text { and }\frac{x}{1} = \frac{y}{2} = \frac{z}{3}\]  are perpendicular to each other. 


Find the angle between the following pair of line: 

\[\overrightarrow{r} = \left( 4 \hat{i} - \hat{j} \right) + \lambda\left( \hat{i} + 2 \hat{j} - 2 \hat{k} \right) \text{ and }\overrightarrow{r} = \hat{i} - \hat{j} + 2 \hat{k} - \mu\left( 2 \hat{i} + 4 \hat{j} - 4 \hat{k} \right)\]


Find the angle between the following pair of line: 

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


Find the angle between the following pair of line: 

\[\overrightarrow{r} = \lambda\left( \hat{i} + \hat{j} + 2 \hat{k} \right) \text{ and } \overrightarrow{r} = 2 \hat{j} + \mu\left\{ \left( \sqrt{3} - 1 \right) \hat{i} - \left( \sqrt{3} + 1 \right) \hat{j} + 4 \hat{k} \right\}\]

 


Find the angle between the pairs of lines with direction ratios proportional to 5, −12, 13 and −3, 4, 5


Find the equation of the line passing through the point (1, 2, −4) and parallel to the line \[\frac{x - 3}{4} = \frac{y - 5}{2} = \frac{z + 1}{3} .\] 


Determine the equations of the line passing through the point (1, 2, −4) and perpendicular to the two lines \[\frac{x - 8}{8} = \frac{y + 9}{- 16} = \frac{z - 10}{7} \text{    and    } \frac{x - 15}{3} = \frac{y - 29}{8} = \frac{z - 5}{- 5}\]


Find the vector equation of the line passing through the point (2, −1, −1) which is parallel to the line 6x − 2 = 3y + 1 = 2z − 2. 


Show that the lines \[\vec{r} = 3 \hat{i} + 2 \hat{j} - 4 \hat{k} + \lambda\left( \hat{i} + 2 \hat{j} + 2 \hat{k} \right) \text{ and } \vec{r} = 5 \hat{i} - 2 \hat{j}  + \mu\left( 3 \hat{i} + 2 \hat{j} + 6 \hat{k} \right)\] are intersecting. Hence, find their point of intersection.


Find the perpendicular distance of the point (3, −1, 11) from the line \[\frac{x}{2} = \frac{y - 2}{- 3} = \frac{z - 3}{4} .\]


Find the length of the perpendicular drawn from the point (5, 4, −1) to the line \[\overrightarrow{r} = \hat{i}  + \lambda\left( 2 \hat{i} + 9 \hat{j} + 5 \hat{k} \right) .\]


Find the equation of the perpendicular drawn from the point P (−1, 3, 2) to the line  \[\overrightarrow{r} = \left( 2 \hat{j} + 3 \hat{k} \right) + \lambda\left( 2 \hat{i} + \hat{j} + 3 \hat{k}  \right) .\]  Also, find the coordinates of the foot of the perpendicular from P.


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


Find the shortest distance between the following pairs of lines whose vector equations are: \[\overrightarrow{r} = \left( 2 \hat{i} - \hat{j} - \hat{k}  \right) + \lambda\left( 2 \hat{i}  - 5 \hat{j} + 2 \hat{k}  \right) \text{ and }, \overrightarrow{r} = \left( \hat{i} + 2 \hat{j} + \hat{k} \right) + \mu\left( \hat{i} - \hat{j}  + \hat{k}  \right)\]


Find the shortest distance between the following pairs of lines whose vector are: \[\overrightarrow{r} = \left( \hat{i} + \hat{j} \right) + \lambda\left( 2 \hat{i} - \hat{j} + \hat{k} \right) \text{ and } , \overrightarrow{r} = 2 \hat{i} + \hat{j} - \hat{k} + \mu\left( 3 \hat{i} - 5 \hat{j} + 2 \hat{k} \right)\]


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


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


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)


The cartesian equations of a line AB are  \[\frac{2x - 1}{\sqrt{3}} = \frac{y + 2}{2} = \frac{z - 3}{3} .\]   Find the direction cosines of a line parallel to AB


If the equations of a line AB are 

\[\frac{3 - x}{1} = \frac{y + 2}{- 2} = \frac{z - 5}{4},\] write the direction ratios of a line parallel to AB


The angle between the straight lines \[\frac{x + 1}{2} = \frac{y - 2}{5} = \frac{z + 3}{4} and \frac{x - 1}{1} = \frac{y + 2}{2} = \frac{z - 3}{- 3}\] is


The perpendicular distance of the point P (1, 2, 3) from the line \[\frac{x - 6}{3} = \frac{y - 7}{2} = \frac{z - 7}{- 2}\] is 

 


The equation of the line passing through the points \[a_1 \hat{i}  + a_2 \hat{j}  + a_3 \hat{k}  \text{ and }  b_1 \hat{i} + b_2 \hat{j}  + b_3 \hat{k} \]  is 


Find the value of p for which the following lines are perpendicular : 

`(1-x)/3 = (2y-14)/(2p) = (z-3)/2 ; (1-x)/(3p) = (y-5)/1 = (6-z)/5`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×