English

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)

Advertisements
Advertisements

Question

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)

Sum
Advertisements

Solution

The equation of the line passing through the points (1, 3, 0) and (0, 3, 0) is

\[\frac{x - 1}{0 - 1} = \frac{y - 3}{3 - 3} = \frac{z - 0}{0 - 0}\]

\[ = \frac{x - 1}{- 1} = \frac{y - 3}{0} = \frac{z}{0}\]


Since the first line passes through the point (0, 0, 0) and has direction ratios proportional to 1, 0, 2, its vector equation is \[\overrightarrow{r} = \overrightarrow{a_1} + \lambda \overrightarrow{b_1} . . . (1) \]

\[\text{ Here }, \]

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

\[ \overrightarrow{b_1} = \hat{i} + 0 \hat{j}+ 2 \hat{k} \]

Also, the second line passes through the point (1, 3, 0) and has direction ratios proportional to -1, 0 , 0 .

Its vector equation is 

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

\[\text{ Here }, \]

\[ \overrightarrow{a_2} = \hat{i} + 3 \hat{j} + 0 \hat{k} \]

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

Now,

\[\overrightarrow{a_2} - \overrightarrow{a_1} = \hat{i} + 3 \hat{j} + 0 \hat{k} \]

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

\[ \Rightarrow \left| \overrightarrow{b_1} \times \overrightarrow{b_2} \right| = \sqrt{0^2 + \left( - 2 \right)^2 + 0^2}\]

\[ = \sqrt{0 + 4 + 0}\]

\[ = 2\]

\[\text{ and }  \left( \overrightarrow{a_2} - \overrightarrow{a_1} \right) . \left( \overrightarrow{b_1} \times \overrightarrow{b_2} \right) = \left( \hat{i} + 3 \hat{j} + 0 \hat{k}  \right) . \left( 0 \hat{i}  - 2 \hat{j}  + 0 \hat{k}  \right)\]

\[ = - 6\]

The shortest distance between the lines

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

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

\[ = \left| \frac{- 6}{2} \right|\]

\[ = 3 \]

\[ \therefore d = \text{ 3 } units\]

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 5.2 | Page 38

RELATED QUESTIONS

The Cartestation equation of  line is `(x-6)/2=(y+4)/7=(z-5)/3` find its vector equation.


Find the vector and Cartesian equations of the line through the point (1, 2, −4) and perpendicular to the two lines. 

`vecr=(8hati-19hatj+10hatk)+lambda(3hati-16hatj+7hatk) " and "vecr=(15hati+29hatj+5hatk)+mu(3hati+8hatj-5hatk)`

 

 


If the Cartesian equations of a line are ` (3-x)/5=(y+4)/7=(2z-6)/4` , write the vector equation for the line.


 

A line passes through (2, −1, 3) and is perpendicular to the lines `vecr=(hati+hatj-hatk)+lambda(2hati-2hatj+hatk) and vecr=(2hati-hatj-3hatk)+mu(hati+2hatj+2hatk)` . Obtain its equation in vector and Cartesian from. 

 

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).


The Cartesian equation of a line is `(x-5)/3 = (y+4)/7 = (z-6)/2` Write its vector form.


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)`


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.


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 line through the points (1, −1, 2) and (3, 4, −2) is perpendicular to the through the points (0, 3, 2) and (3, 5, 6).


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 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 angle between the pairs of lines with direction ratios proportional to  2, 2, 1 and 4, 1, 8 .

 


Find the angle between the pairs of lines with direction ratios proportional to   abc and b − cc − aa − b.


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


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


If the lines \[\frac{x - 1}{- 3} = \frac{y - 2}{2 \lambda} = \frac{z - 3}{2} \text{     and     } \frac{x - 1}{3\lambda} = \frac{y - 1}{1} = \frac{z - 6}{- 5}\]  are perpendicular, find the value of λ.


If the coordinates of the points ABCD be (1, 2, 3), (4, 5, 7), (−4, 3, −6) and (2, 9, 2) respectively, then find the angle between the lines AB and CD


Find the value of λ so that the following lines are perpendicular to each other. \[\frac{x - 5}{5\lambda + 2} = \frac{2 - y}{5} = \frac{1 - z}{- 1}, \frac{x}{1} = \frac{2y + 1}{4\lambda} = \frac{1 - z}{- 3}\]


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.  


Determine whether the following pair 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 foot of the perpendicular drawn from the point A (1, 0, 3) to the joint of the points B (4, 7, 1) and C (3, 5, 3). 


Find the shortest distance between the following pairs of lines whose vector equations are: \[\overrightarrow{r} = \left( 3 \hat{i} + 5 \hat{j} + 7 \hat{k} \right) + \lambda\left( \hat{i} - 2 \hat{j} + 7 \hat{k} \right) \text{ and } \overrightarrow{r} = - \hat{i} - \hat{j} - \hat{k}  + \mu\left( 7 \hat{i}  - 6 \hat{j}  + \hat{k}  \right)\]


Write the cartesian and vector equations of Z-axis.

 

Write the direction cosines of the line \[\frac{x - 2}{2} = \frac{2y - 5}{- 3}, z = 2 .\]


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


Write the angle between the lines 2x = 3y = −z and 6x = −y = −4z.

 

Write the vector equation of a line given by \[\frac{x - 5}{3} = \frac{y + 4}{7} = \frac{z - 6}{2} .\]

 


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 straight line \[\frac{x - 3}{3} = \frac{y - 2}{1} = \frac{z - 1}{0}\] is


If 2x + y = 0 is one of the line represented by 3x2 + kxy + 2y2 = 0 then k = ______ 


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 vector equation of the lines passing through the point having position vector `(-hati - hatj + 2hatk)` and parallel to the line `vecr = (hati + 2hatj + 3hatk) + λ(3hati + 2hatj + hatk)`.


A line passes through the point (2, – 1, 3) and is perpendicular to the lines `vecr = (hati + hatj - hatk) + λ(2hati - 2hatj + hatk)` and `vecr = (2hati - hatj - 3hatk) + μ(hati + 2hatj + 2hatk)` obtain its equation.


Which parametric form corresponds to a line through \[A(x_1,y_1,z_1)\] with direction ratios \[a,b,c\]?


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×