हिंदी

Show that the Line Through the Points (4, 7, 8) and (2, 3, 4) is Parallel to the Line Through the Points (−1, −2, 1) And, (1, 2, 5). - Mathematics

Advertisements
Advertisements

प्रश्न

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

योग

उत्तर

Equations of lines passing through the points 

\[\left( x_1 , y_1 , z_1 \right) \text { and } \left  ( x_2 , y_2 , z_2 \right)\] are given by 

\[\frac{x - x_1}{x_2 - x_1} = \frac{y - y_1}{y_2 - y_1} = \frac{z - z_1}{z_2 - z_1}\]

So, the equation of a line passing through (4, 7, 8) and (2, 3, 4) is

\[\frac{x - 4}{2 - 4} = \frac{y - 7}{3 - 7} = \frac{z - 8}{4 - 8}\]

\[ \Rightarrow \frac{x - 4}{- 2} = \frac{y - 7}{- 4} = \frac{z - 8}{- 4}\]

Also, the equation of the line passing through the points ( -1,-2 ,1) and (1,2,5) is 

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

\[ \Rightarrow \frac{x + 1}{2} = \frac{y + 2}{4} = \frac{z - 1}{4}\]

We know that two lines are parallel if

\[\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \]

\[\text { Cartesian equations of the two lines are given by } \]

\[\frac{x - x_1}{a_1} = \frac{y - y_1}{b_1} = \frac{z - z_1}{c_1} \text { and } \frac{x - x_2}{a_2} = \frac{y - y_2}{b_2} = \frac{z - z_2}{c_2}\]

We observe 

\[\frac{- 2}{2} = \frac{- 4}{4} = \frac{- 4}{4} = - 1\]

Hence, the given lines are parallel to each other.

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 28: Straight Line in Space - Exercise 28.2 [पृष्ठ १६]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
अध्याय 28 Straight Line in Space
Exercise 28.2 | Q 3 | पृष्ठ १६

वीडियो ट्यूटोरियलVIEW ALL [4]

संबंधित प्रश्न

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

 


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


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)`

 

 


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


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 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.


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


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


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


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 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 pairs of lines with direction ratios proportional to  1, 2, −2 and −2, 2, 1 .


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


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. 


Determine whether the following pair of lines intersect or not: 

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


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


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 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 equation of the perpendicular drawn from the point P (2, 4, −1) to the line  \[\frac{x + 5}{1} = \frac{y + 3}{4} = \frac{z - 6}{- 9} .\]  Also, write down the coordinates of the foot of the perpendicular from P


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


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


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


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


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


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


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

 

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


Write the formula for the shortest distance between the lines 

\[\overrightarrow{r} = \overrightarrow{a_1} + \lambda \overrightarrow{b} \text{ and }  \overrightarrow{r} = \overrightarrow{a_2} + \mu \overrightarrow{b} .\] 

 


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


The angle between the lines

\[\frac{x - 1}{1} = \frac{y - 1}{1} = \frac{z - 1}{2} \text{ and }, \frac{x - 1}{- \sqrt{3} - 1} = \frac{y - 1}{\sqrt{3} - 1} = \frac{z - 1}{4}\] is 

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


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

 

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×