English

Write the Cartesian and Vector Equations of Z-axis. - Mathematics

Advertisements
Advertisements

Question

Write the cartesian and vector equations of Z-axis.

 
Short/Brief Note
Advertisements

Solution

Since z-axis passes through the the point (0, 0, 0) having position vector  \[\overrightarrow{a} = 0 \hat{i} + 0 \hat{j} + 0 \hat{k}\] and is parallel to the vector \[\overrightarrow{b} = 0 \hat{i} + 0 \hat{j} + \hat{k}\] having direction ratios proportional to 0, 0, 1, the cartesian equation of z-axis is 

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

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

Also, its vector equation is , 

\[\overrightarrow{r} = \overrightarrow{a} + \lambda \overrightarrow{b} \]

\[ = 0 \hat{i} + 0 \hat{j} + 0 \hat{k} + \lambda\left( 0 \hat{i} + 0 \hat{j} + \hat{k} \right)\]

\[ = \lambda \hat{k}  \]

 

shaalaa.com
  Is there an error in this question or solution?
Chapter 28: Straight Line in Space - Very Short Answers [Page 41]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 28 Straight Line in Space
Very Short Answers | Q 3 | Page 41

RELATED QUESTIONS

 

Find the value of p, so that the lines `l_1:(1-x)/3=(7y-14)/p=(z-3)/2 and l_2=(7-7x)/3p=(y-5)/1=(6-z)/5 ` are perpendicular to each other. Also find the equations of a line passing through a point (3, 2, – 4) and parallel to line l1.

 

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`


Find the equation of a line parallel to x-axis and passing through the origin.


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 the line passing through the points (−1, 0, 2) and (3, 4, 6).


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).


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.


Show that the points whose position vectors are  \[- 2 \hat{i} + 3 \hat{j} , \hat{i} + 2 \hat{j} + 3 \hat{k}  \text{ and }  7 \text{ i}  - \text{ k} \]  are collinear.


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


Find the angle between the following pair of line:

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


Find the angle between the following pair of line:

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


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


Show that the lines \[\frac{x - 5}{7} = \frac{y + 2}{- 5} = \frac{z}{1} \text{ and } \frac{x}{1} = \frac{y}{2} = \frac{z}{3}\] are perpendicular to each other.


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.


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. 


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 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 foot of the perpendicular from (1, 2, −3) to the line \[\frac{x + 1}{2} = \frac{y - 3}{- 2} = \frac{z}{- 1} .\]


Find the shortest distance between the following pairs of lines whose cartesian equations are: \[\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} and \frac{x - 2}{3} = \frac{y - 3}{4} = \frac{z - 5}{5}\] 


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 following pairs of parallel lines whose equations are: \[\overrightarrow{r} = \left( \hat{i} + \hat{j} \right) + \lambda\left( 2 \hat{i} - \hat{j} + \hat{k} \right) \text{ and } \overrightarrow{r} = \left( 2 \hat{i} + \hat{j} - \hat{k} \right) + \mu\left( 4 \hat{i} - 2 \hat{j} + 2 \hat{k} \right)\]


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) 


Find the shortest distance between the lines \[\frac{x + 1}{7} = \frac{y + 1}{- 6} = \frac{z + 1}{1} \text{ and }  \frac{x - 3}{1} = \frac{y - 5}{- 2} = \frac{z - 7}{1}\]


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

 

 


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


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

 


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


If the direction ratios of a line are proportional to 1, −3, 2, then its direction cosines are

 


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.


Choose correct alternatives:

The difference between the slopes of the lines represented by 3x2 - 4xy + y2 = 0 is 2


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`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×