English

Find-shortest-distance-between-pairs-lines-whose-vector-are-vec-r-left-8-3-lambda-right-hat-i-left-9-16-lambda-right-hat-j-left-10-7-lambda-right-hat-k - Mathematics

Advertisements
Advertisements

Question

Find the shortest distance between the following pairs of lines whose vector equations are: \[\overrightarrow{r} = \left( 8 + 3\lambda \right) \hat{i} - \left( 9 + 16\lambda \right) \hat{j} + \left( 10 + 7\lambda \right) \hat{k} \]\[\overrightarrow{r} = 15 \hat{i} + 29 \hat{j} + 5 \hat{k} + \mu\left( 3 \hat{i}  + 8 \hat{j} - 5 \hat{k}  \right)\]

Sum

Solution

The vector equations of the given lines can be re-written as 

\[\overrightarrow{r} = 8 \hat{i} - 9 \hat{j} + 10 \hat{k}  + \lambda\left( 3 \hat{i}  - 16 \hat{j}  + 7 \hat{k}  \right)\] and \[\overrightarrow{r} = 15 \hat{i} + 29 \hat{j} + 5 \hat{k} + \mu\left( 3 \hat{i} + 8 \hat{j} - 5 \hat{k} \right)\]

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 , 

\[\overrightarrow{a_1} = 8 \hat{i} - 9 \hat{j} + 10 \hat{k}\]

\[\overrightarrow{b_1} = 3 \hat{i} - 16 \hat{j} + 7 \hat{k}\]

\[\overrightarrow{a_2} = 15 \hat{i} + 29 \hat{j} + 5 \hat{k}\]

\[\overrightarrow{b_2} = 3 \hat{i} + 8 \hat{j} - 5 \hat{k}\]

\[\therefore \overrightarrow{a_2} - \overrightarrow{a_1} = \left( 15 \hat{i} + 29 \hat{j} + 5 \hat{k} \right) - \left( 8 \hat{i} - 9 \hat{j} + 10 \hat{k} \right) = 7 \hat{i} + 38 \hat{j} - 5 \hat{k} \]

\[\overrightarrow{b_1} \times \vec{b_2} = \begin{vmatrix}\hat{i}  & \hat{j} & \hat{k} \\ 3 & - 16 & 7 \\ 3 & 8 & - 5\end{vmatrix} = 24 \hat{i} + 36 \hat{j} + 72 \hat{k} \]

\[ \Rightarrow \left| \overrightarrow{b_1} \times \overrightarrow{b_2} \right| = \sqrt{{24}^2 + {36}^2 + {72}^2} = \sqrt{576 + 1296 + 5184} = \sqrt{7056} = 84\]

Also,

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

\[ = \left( 7 \hat{i} + 38 \hat{j} - 5 \hat{k} \right) . \left( 24 \hat{i} + 36 \hat{j} + 72 \hat{k} \right)\]

\[ = 7 \times 24 + 38 \times 36 + \left( - 5 \right) \times 72\]

\[ = 168 + 1368 - 360\]

\[ = 1176\]

We know that the shortest distance between the lines

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

\[d = \left| \frac{\left( \overrightarrow{a_2} - \overrightarrow{a_1} \right) . \left( \overrightarrow{b_1} \times \overrightarrow{b_2} \right)}{\left| \overrightarrow{b_1} \times \overrightarrow{b_2} \right|} \right|\]

∴ Required shortest distance between the given pairs of lines,

\[d = \left| \frac{\left( \overrightarrow{a_2} - \overrightarrow{a_1} \right) . \left( \overrightarrow{b_1} \times \overrightarrow{b_2} \right)}{\left| \overrightarrow{b_1} \times \overrightarrow{b_2} \right|} \right|\]

\[ = \left| \frac{1176}{84} \right|\]

\[ = 14\]

shaalaa.com
  Is there an error in this question or solution?
Chapter 28: Straight Line in Space - Exercise 28.5 [Page 37]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 28 Straight Line in Space
Exercise 28.5 | Q 1.8 | Page 37

RELATED QUESTIONS

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


Find the separate equations of the lines represented by the equation 3x2 – 10xy – 8y2 = 0.


Show that the three lines with direction cosines `12/13, (-3)/13, (-4)/13;  4/13, 12/13, 3/13;  3/13, (-4)/13, 12/13 ` are mutually perpendicular.


Show that the line joining the origin to the point (2, 1, 1) is perpendicular to the line determined by the points (3, 5, – 1), (4, 3, – 1).


Find the vector and cartesian equations of the line through the point (5, 2, −4) and which is parallel to the vector  \[3 \hat{i} + 2 \hat{j} - 8 \hat{k} .\]


Find the vector equation of the line passing through the points (−1, 0, 2) and (3, 4, 6).


A line passes through the point with position vector \[2 \hat{i} - 3 \hat{j} + 4 \hat{k} \] and is in the direction of  \[3 \hat{i} + 4 \hat{j} - 5 \hat{k} .\] Find equations of the line in vector and cartesian form. 


ABCD is a parallelogram. The position vectors of the points AB and C are respectively, \[4 \hat{ i} + 5 \hat{j} -10 \hat{k} , 2 \hat{i} - 3 \hat{j} + 4 \hat{k}  \text{ and } - \hat{i} + 2 \hat{j} + \hat{k} .\]  Find the vector equation of the line BD. Also, reduce it to cartesian form.


Find in vector form as well as in cartesian form, the equation of the line passing through the points A (1, 2, −1) and B (2, 1, 1).


Find the vector equation of a line passing through (2, −1, 1) and parallel to the line whose equations are \[\frac{x - 3}{2} = \frac{y + 1}{7} = \frac{z - 2}{- 3} .\]


The cartesian equations of a line are \[\frac{x - 5}{3} = \frac{y + 4}{7} = \frac{z - 6}{2} .\]  Find a vector equation for the line.


Find the direction cosines of the line  \[\frac{4 - x}{2} = \frac{y}{6} = \frac{1 - z}{3} .\]  Also, reduce it to vector form. 


Find the angle between the following pair of line:

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


Find the angle between the pairs of lines with direction ratios proportional to  1, 2, −2 and −2, 2, 1 .


Find the equation of the line passing through the point (1, −1, 1) and perpendicular to the lines joining the points (4, 3, 2), (1, −1, 0) and (1, 2, −1), (2, 1, 1).


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 direction cosines of the line 

\[\frac{x + 2}{2} = \frac{2y - 7}{6} = \frac{5 - z}{6}\]  Also, find the vector equation of the line through the point A(−1, 2, 3) and parallel to the given line.  


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 line passing through the points A (0, 6, −9) and B (−3, −6, 3). If D is the foot of perpendicular drawn from a point C (7, 4, −1) on the line AB, then find the coordinates of the point D and the equation of line CD


Find the distance of the point (2, 4, −1) from the line  \[\frac{x + 5}{1} = \frac{y + 3}{4} = \frac{z - 6}{- 9}\] 


Find the shortest distance between the following pairs of lines whose vector equations are: \[\vec{r} = 3 \hat{i} + 8 \hat{j} + 3 \hat{k}  + \lambda\left( 3 \hat{i}  - \hat{j}  + \hat{k}  \right) \text{ and }  \vec{r} = - 3 \hat{i}  - 7 \hat{j}  + 6 \hat{k}  + \mu\left( - 3 \hat{i}  + 2 \hat{j}  + 4 \hat{k} \right)\]


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


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


Find the shortest distance between the following pairs of lines whose cartesian equations are:  \[\frac{x - 3}{1} = \frac{y - 5}{- 2} = \frac{z - 7}{1} \text{ and } \frac{x + 1}{7} = \frac{y + 1}{- 6} = \frac{z + 1}{1}\]


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


Write the vector equations of the following lines and hence determine the distance between them  \[\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z + 4}{6} \text{ and } \frac{x - 3}{4} = \frac{y - 3}{6} = \frac{z + 5}{12}\]


Write the direction cosines of the line whose cartesian equations are 6x − 2 = 3y + 1 = 2z − 4.

 

The equations of a line are given by \[\frac{4 - x}{3} = \frac{y + 3}{3} = \frac{z + 2}{6} .\]  Write the direction cosines of a line parallel to this line.


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 direction ratios of the line x − y + z − 5 = 0 = x − 3y − 6 are proportional to

 

 


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 


The separate equations of the lines represented by `3x^2 - 2sqrt(3)xy - 3y^2` = 0 are ______ 


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.


Find the equations of the diagonals of the parallelogram PQRS whose vertices are P(4, 2, – 6), Q(5, – 3, 1), R(12, 4, 5) and S(11, 9, – 2). Use these equations to find the point of intersection of diagonals.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×