English

Find the Equation of the Line Passing Through the Point (2, −1, 3) and Parallel to the Line → R = ( ^ I − 2 ^ J + ^ K ) + λ ( 2 ^ I + 3 ^ J − 5 ^ K ) .

Advertisements
Advertisements

Question

Find the equation of the line passing through the point (2, −1, 3) and parallel to the line  \[\overrightarrow{r} = \left( \hat{i} - 2 \hat{j} + \hat{k} \right) + \lambda\left( 2 \hat{i} + 3 \hat{j} - 5 \hat{k} \right) .\]

Sum
Advertisements

Solution

The given line is parallel to the vector  \[2 \hat{i} + 3 \hat{j} - 5 \hat{k} \] and the required line is parallel to the given line. So, the required line is parallel to the vector \[2 \hat{i} + 3 \hat{j} - 5 \hat{k} \]  Hence, the equation of the required line passing through the point (2,-1, 3) and parallel to the vector  \[2 \hat{i} + 3 \hat{j} - 5 \hat{k} \]  is  \[\overrightarrow{r} = \left( 2 \hat{i} - \hat{j} + 3 \hat{k} \right) + \lambda\left( 2 \hat{i} + 3 \hat{j} - 5 \hat{k} \right)\] 

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 14 | Page 16

RELATED QUESTIONS

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

 


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`


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.


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


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 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.


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 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.


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 cartesian equation of a line passing through (1, −1, 2) and parallel to the line whose equations are  \[\frac{x - 3}{1} = \frac{y - 1}{2} = \frac{z + 1}{- 2}\]  Also, reduce the equation obtained in 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 following pair of line: 

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


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


Find the equations of the line passing through the point (2, 1, 3) and perpendicular to the lines  \[\frac{x - 1}{1} = \frac{y - 2}{2} = \frac{z - 3}{3} \text{  and  } \frac{x}{- 3} = \frac{y}{2} = \frac{z}{5}\]


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 A, B, C, D 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 equation of the perpendicular drawn from the point P (−1, 3, 2) to the line  \[\overrightarrow{r} = \left( 2 \hat{j} + 3 \hat{k} \right) + \lambda\left( 2 \hat{i} + \hat{j} + 3 \hat{k}  \right) .\]  Also, find the coordinates of the foot of the perpendicular from P.


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:  \[\frac{x - 3}{1} = \frac{y - 5}{- 2} = \frac{z - 7}{1} \text{ and } \frac{x + 1}{7} = \frac{y + 1}{- 6} = \frac{z + 1}{1}\]


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 equations of the lines joining the following pairs of vertices and then find the shortest distance between the lines
(i) (0, 0, 0) and (1, 0, 2) 


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 direction cosines of the line whose cartesian equations are 2x = 3y = −z.

 

The angle between the straight lines \[\frac{x + 1}{2} = \frac{y - 2}{5} = \frac{z + 3}{4} and \frac{x - 1}{1} = \frac{y + 2}{2} = \frac{z - 3}{- 3}\] is


The direction ratios of the line perpendicular to the lines \[\frac{x - 7}{2} = \frac{y + 17}{- 3} = \frac{z - 6}{1} \text{ and }, \frac{x + 5}{1} = \frac{y + 3}{2} = \frac{z - 4}{- 2}\] are proportional to


The lines  \[\frac{x}{1} = \frac{y}{2} = \frac{z}{3} \text { and } \frac{x - 1}{- 2} = \frac{y - 2}{- 4} = \frac{z - 3}{- 6}\] 

 


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


If y – 2x – k = 0 touches the conic 3x2 – 5y2 = 15, find the value of k. 


Find the joint equation of pair of lines through the origin which is perpendicular to the lines represented by 5x2 + 2xy - 3y2 = 0 


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`


For a line passing through a point with position vector \[\vec{a}\] and parallel to a vector \[\vec{b}\], which is the vector equation?


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


What is the parametric form of the line through \[(5,2,-4)\] with direction ratios \[3,2,-8\]?


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×