मराठी

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

Advertisements
Advertisements

प्रश्न

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

बेरीज
Advertisements

उत्तर

The direction cosines of the x-axis are 1, 0, 0.

∴ The direction cosines of the line parallel to the x-axis will also be 1, 0, 0.

Hence, the equation of the line through the origin with direction cosines 1, 0, 0 is.

⇒ `(x - 0)/1 = (y - 0)/0 = (z - 0)/0`

⇒ `x/1 = y/0 = z/0`

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 11: Three Dimensional Geometry - Exercise 11.4 [पृष्ठ ४९८]

APPEARS IN

एनसीईआरटी Mathematics Part 1 and 2 [English] Class 12
पाठ 11 Three Dimensional Geometry
Exercise 11.4 | Q 4 | पृष्ठ ४९८

व्हिडिओ 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 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)`

 

 


If the Cartesian equations of a line are ` (3-x)/5=(y+4)/7=(2z-6)/4` , write the vector equation for the line.


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 the line which passes through the point (1, 2, 3) and is parallel to the vector `3hati+2hatj-2hatk`.


Find the vector and Cartesian equations of a line passing through (1, 2, –4) and perpendicular to the two lines `(x - 8)/3 = (y + 19)/(-16) = (z - 10)/7` and `(x - 15)/3 = (y - 29)/8 = (z - 5)/(-5)`


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


Find the points on the line \[\frac{x + 2}{3} = \frac{y + 1}{2} = \frac{z - 3}{2}\]  at a distance of 5 units from the point P (1, 3, 3).


Show that the three lines with direction cosines \[\frac{12}{13}, \frac{- 3}{13}, \frac{- 4}{13}; \frac{4}{13}, \frac{12}{13}, \frac{3}{13}; \frac{3}{13}, \frac{- 4}{13}, \frac{12}{13}\] are mutually perpendicular. 


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


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 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:

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


Find the angle between the following pair of line:

\[\frac{x - 5}{1} = \frac{2y + 6}{- 2} = \frac{z - 3}{1} \text{  and  } \frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 6}{5}\]


Find the angle between two lines, one of which has direction ratios 2, 2, 1 while the  other one is obtained by joining the points (3, 1, 4) and (7, 2, 12). 


Find the vector equation of the line passing through the point (2, −1, −1) which is parallel to the line 6x − 2 = 3y + 1 = 2z − 2. 


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.


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


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


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


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 value of λ for which the lines  \[\frac{x - 3}{- 3} = \frac{y + 2}{2\lambda} = \frac{z + 4}{2} \text{ and } \frac{x + 1}{3\lambda} = \frac{y - 2}{1} = \frac{z + 6}{- 5}\]  are perpendicular to each other.


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


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

 


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.


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


The distance of the point (4, 3, 8) from the Y-axis is ______.


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`


P is a point on the line joining the points A(0, 5, −2) and B(3, −1, 2). If the x-coordinate of P is 6, then its z-coordinate 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×