English

Find the Direction Cosines of the Line X + 2 2 = 2 Y − 7 6 = 5 − Z 6 Also, Find the Vector Equation of the Line Through the Point A(−1, 2, 3) and Parallel to the Given Line. - Mathematics

Advertisements
Advertisements

Question

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.  

Sum
Advertisements

Solution

The equation of the given line is 

\[\frac{x + 2}{2} = \frac{2y - 7}{6} = \frac{5 - z}{6}\]

The given equation can be re-written as

\[\frac{x + 2}{2} = \frac{y - \frac{7}{2}}{3} = \frac{z - 5}{- 6}\]

This line passes through the point \[\left( - 2, \frac{7}{2}, 5 \right)\] and has direction ratios proportional to 2, 3, −6. So, its direction cosines are  \[\frac{2}{\sqrt{2^2 + 3^2 + \left( - 6 \right)^2}}, \frac{3}{\sqrt{2^2 + 3^2 + \left( - 6 \right)^2}}, \frac{- 6}{\sqrt{2^2 + 3^2 + \left( - 6 \right)^2}}\]  Or \[\frac{2}{7}, \frac{3}{7}, \frac{- 6}{7}\] The required line passes through the point having position vector  \[\overrightarrow{a} = - \hat{i} + 2 \hat{j} + 3 \hat{k}\] and is parallel to the vector \[\overrightarrow{b} = 2 \hat{i} + 3 \hat{j} - 6 \hat{k}\]  So, its vector equation is  \[\overrightarrow{r} = \left( - \hat{i}  + 2 \hat{j} + 3 \hat{k} \right) + \lambda\left( 2 \hat{i}  + 3 \hat{j}  - 6 \hat{k}  \right)\]

shaalaa.com
  Is there an error in this question or solution?
Chapter 28: Straight Line in Space - Exercise 28.2 [Page 17]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 28 Straight Line in Space
Exercise 28.2 | Q 24 | Page 17

RELATED QUESTIONS

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 equation of the lines which passes through the point with position vector `4hati - hatj +2hatk` and is in the direction of `-2hati + hatj + hatk`


Find the vector equation of a line which is parallel to the vector \[2 \hat{i} - \hat{j} + 3 \hat{k}\]  and which passes through the point (5, −2, 4). Also, reduce it to cartesian form.


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


Find the vector equation of the line passing through the point A(1, 2, –1) and parallel to the line 5x – 25 = 14 – 7y = 35z.


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).


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 following pair of line:

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


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.


Show that the lines  \[\frac{x}{1} = \frac{y - 2}{2} = \frac{z + 3}{3} \text{          and         } \frac{x - 2}{2} = \frac{y - 6}{3} = \frac{z - 3}{4}\] intersect and find their point of intersection. 


Prove that the lines through A (0, −1, −1) and B (4, 5, 1) intersects the line through C (3, 9, 4) and D (−4, 4, 4). Also, find their point of intersection. 


Prove that the line \[\vec{r} = \left( \hat{i }+ \hat{j }- \hat{k} \right) + \lambda\left( 3 \hat{i} - \hat{j} \right) \text{ and } \vec{r} = \left( 4 \hat{i} - \hat{k} \right) + \mu\left( 2 \hat{i} + 3 \hat{k} \right)\] intersect and find their point of intersection.


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 foot of the perpendicular drawn from the point  \[\hat{i} + 6 \hat{j} + 3 \hat{k} \]  to the line  \[\overrightarrow{r} = \hat{j} + 2 \hat{k} + \lambda\left( \hat{i} + 2 \hat{j} + 3 \hat{k}  \right) .\]  Also, find the length of the perpendicular


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


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


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


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.

 

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


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

 

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

 


Find the Cartesian equations of the line which passes through the point (−2, 4 , −5) and is parallel to the line \[\frac{x + 3}{3} = \frac{4 - y}{5} = \frac{z + 8}{6} .\]


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 \[\vec{r} = 7 \hat{i} - 6 \hat{k}  + \mu\left( \hat{i}  + 2 \hat{j}  + 2 \hat{k}  \right)\] 


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


The shortest distance between the lines  \[\frac{x - 3}{3} = \frac{y - 8}{- 1} = \frac{z - 3}{1} \text{ and }, \frac{x + 3}{- 3} = \frac{y + 7}{2} = \frac{z - 6}{4}\] 

 

 

 

 


 The equation of a line is 2x -2 = 3y +1 = 6z -2 find the direction ratios and also find the vector equation of the line. 


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


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.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×