English

Find the Angle Between the Following Pairs of Lines: − X + 2 − 2 = Y − 1 7 = Z + 3 − 3 a N D X + 2 − 1 = 2 Y − 8 4 = Z − 5 4

Advertisements
Advertisements

Question

Find the angle between the following pair of line:

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

Sum
Advertisements

Solution

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

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

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

Let

\[\overrightarrow{b_1}\] and \[\overrightarrow{b_2}\] be vectors parallel to the given lines.

Now,

\[\overrightarrow{b_1} = 2 \hat{i} + 7 \hat{j} - 3 \hat{k} \]

\[ \overrightarrow{b_2} = - 1 \hat{i} + 2 \hat{j} + 4 \hat{k}\]

If θ   is the angle between the given lines, then

\[\cos \theta = \frac{\overrightarrow{b_1} . \overrightarrow{b_2}}{\left| \overrightarrow{b_1} \right| \left| \overrightarrow{b_2} \right|}\]

\[ = \frac{\left( 2 \hat{i} + 7 \hat{j} - 3 \hat{k} \right) . \left( - 1 \hat{i} + 2 \hat{j} + 4 \hat{k} \right)}{\sqrt{2^2 + 7^2 + \left( - 3 \right)^2} \sqrt{\left( - 1 \right)^2 + 2^2 + 4^2}}\]

\[ = \frac{- 2 + 14 - 12}{\sqrt{62} \sqrt{21}}\]

\[ = 0\]

\[ \Rightarrow \theta = \frac{\pi}{2}\]

shaalaa.com
  Is there an error in this question or solution?
Chapter 27: Straight Line in Space - Exercise 28.2 [Page 16]

APPEARS IN

R.D. Sharma Mathematics Volume 1 and 2 [English] Class 12
Chapter 27 Straight Line in Space
Exercise 28.2 | Q 9.6 | Page 16

RELATED QUESTIONS

The Cartesian equations of line are 3x+1=6y-2=1-z find its equation in vector form.

 


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


Show that the lines `(x-5)/7 = (y + 2)/(-5) = z/1` and `x/1 = y/2 = z/3` are perpendicular to each other.


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 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 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 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.  


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.


Determine whether the following pair 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 perpendicular distance of the point (1, 0, 0) from the line  \[\frac{x - 1}{2} = \frac{y + 1}{- 3} = \frac{z + 10}{8}.\] Also, find the coordinates of the foot of the perpendicular and the equation of the perpendicular.


Find the foot of perpendicular from the point (2, 3, 4) to the line \[\frac{4 - x}{2} = \frac{y}{6} = \frac{1 - z}{3} .\] Also, find the perpendicular distance from the given point to the 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 shortest distance between the following pairs of lines whose vector equations are: \[\overrightarrow{r} = \left( 3 \hat{i} + 5 \hat{j} + 7 \hat{k} \right) + \lambda\left( \hat{i} - 2 \hat{j} + 7 \hat{k} \right) \text{ and } \overrightarrow{r} = - \hat{i} - \hat{j} - \hat{k}  + \mu\left( 7 \hat{i}  - 6 \hat{j}  + \hat{k}  \right)\]


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


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


By computing the shortest distance determine whether the following pairs of lines intersect or not: \[\frac{x - 5}{4} = \frac{y - 7}{- 5} = \frac{z + 3}{- 5} \text{ and } \frac{x - 8}{7} = \frac{y - 7}{1} = \frac{z - 5}{3}\]


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 distance between the lines l1 and l2 given by  \[\overrightarrow{r} = \hat{i} + 2 \hat{j} - 4 \hat{k} + \lambda\left( 2 \hat{i}  + 3 \hat{j}  + 6 \hat{k}  \right) \text{ and } , \overrightarrow{r} = 3 \hat{i} + 3 \hat{j}  - 5 \hat{k}  + \mu\left( 2 \hat{i} + 3 \hat{j}  + 6 \hat{k}  \right)\]

 

 


Write the cartesian and vector equations of X-axis.

 

Cartesian equations of a line AB are  \[\frac{2x - 1}{2} = \frac{4 - y}{7} = \frac{z + 1}{2} .\]   Write the direction ratios of a line parallel to AB.


Write the angle between the lines \[\frac{x - 5}{7} = \frac{y + 2}{- 5} = \frac{z - 2}{1} \text{ and } \frac{x - 1}{1} = \frac{y}{2} = \frac{z - 1}{3} .\]


Find the angle between the lines 2x=3y=-z and 6x =-y=-4z.

 


The lines `x/1 = y/2 = z/3 and (x - 1)/-2 = (y - 2)/-4 = (z - 3)/-6` are


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

 

 


If a line makes angle \[\frac{\pi}{3} \text{ and } \frac{\pi}{4}\]  with x-axis and y-axis respectively, then the angle made by the line with z-axis is


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


Find the equation of a plane which passes through the point (3, 2, 0) and contains the line \[\frac{x - 3}{1} = \frac{y - 6}{5} = \frac{z - 4}{4}\].

 

Show that the lines \[\frac{5 - x}{- 4} = \frac{y - 7}{4} = \frac{z + 3}{- 5} \text { and } \frac{x - 8}{7} = \frac{2y - 8}{2} = \frac{z - 5}{3}\] are coplanar.


Find the value of  λ for which the following lines are perpendicular to each other: 

`(x - 5)/(5 lambda + 2 ) = ( 2 - y )/5 = (1 - z ) /-1 ; x /1 = ( y + 1/2)/(2 lambda ) = ( z -1 ) / 3`


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


The equation of line passing through (3, -1, 2) and perpendicular to the lines `overline("r")=(hat"i"+hat"j"-hat"k")+lambda(2hat"i"-2hat"j"+hat"k")` and `overline("r")=(2hat"i"+hat"j"-3hat"k")+mu(hat"i"-2hat"j"+2hat"k")` is ______.


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


Equation of a line passing through (1, 1, 1) and parallel to z-axis is ______.


In the vector equation \[\vec{r}=\vec{a}+\lambda\vec{b}\], what does \[\vec{r}\] represent?


For the point \[(5,2,-4)\], what is its position vector \[\vec{a}\]?


For a line parallel to \[3\hat{i}+2\hat{j}-8\hat{k}\], what are the direction ratios \[a,b,c\]?


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×