हिंदी

Find the equation of the line in vector and in Cartesian form that passes through the point with position vector 2i^-j^+4k^ and is in the direction i^ +2j^-k^. - Mathematics

Advertisements
Advertisements

प्रश्न

Find the equation of the line in vector and in Cartesian form that passes through the point with position vector `2hati -hatj+4hatk`  and is in the direction `hati + 2hatj - hatk`.

योग
Advertisements

उत्तर

The required line passes through the given vector point `vec(r_1) = 2hati - hatj + 4hatk` and is parallel to the vector `vecb = hati + 2hatj - hatk`.

∴ Equation of required line `vecr = vec(r_1) + λ vecb`

or `vecr = (2hati - hatj + 4hatk) + λ(hati + 2hatj + hatk)`    ........(i)

Cartesian Equations :

Taking `vecr = xhati + yhatj + zhatk` in equation (i),

`xhati + yhatj + zhatk = (2hati - hatj + 4hatk) + λ (hati + 2hatj - hatk)`

⇒ `xhati + yhatj + zhatk = (2 + λ)hati + (-1 + 2 λ)hatj + (4 -  λ)hatk`

⇒ x = 2 + λ, y = −1 + 2λ, z = 4 − λ

⇒ `(x- 2)/1 = (y + 1)/2 = (z - 4)/-1 = λ`

∴ `(x- 2)/1 = (y + 1)/2 = (z - 4)/-1` is the cartesian equation of the line.

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 11: Three Dimensional Geometry - Exercise 11.1 [पृष्ठ ४७७]

APPEARS IN

एनसीईआरटी Mathematics Part 1 and 2 [English] Class 12
अध्याय 11 Three Dimensional Geometry
Exercise 11.1 | Q 5 | पृष्ठ ४७७

वीडियो ट्यूटोरियलVIEW ALL [4]

संबंधित प्रश्न

If a line drawn from the point A( 1, 2, 1) is perpendicular to the line joining P(1, 4, 6) and Q(5, 4, 4) then find the co-ordinates of the foot of the perpendicular.


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


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)`

 

 


 

A line passes through (2, −1, 3) and is perpendicular to the lines `vecr=(hati+hatj-hatk)+lambda(2hati-2hatj+hatk) and vecr=(2hati-hatj-3hatk)+mu(hati+2hatj+2hatk)` . Obtain its equation in vector and Cartesian from. 

 

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.


Find the vector and the Cartesian equations of the line that passes through the points (3, −2, −5), (3, −2, 6).

 


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


Find the vector and cartesian equations of the line through the point (5, 2, −4) and which is parallel to the vector  \[3 \hat{i} + 2 \hat{j} - 8 \hat{k} .\]


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 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 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).


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. 


Find the angle between the following pair of line:

\[\frac{- x + 2}{- 2} = \frac{y - 1}{7} = \frac{z + 3}{- 3} \text{  and  } \frac{x + 2}{- 1} = \frac{2y - 8}{4} = \frac{z - 5}{4}\]


Find the angle between the pairs of lines with direction ratios proportional to   abc and b − cc − aa − b.


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. 


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. 


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.


Determine whether the following pair of lines intersect or not: 

\[\frac{x - 1}{3} = \frac{y - 1}{- 1} = \frac{z + 1}{0} and \frac{x - 4}{2} = \frac{y - 0}{0} = \frac{z + 1}{3}\]


Determine whether the following pair of lines intersect or not:  

\[\frac{x - 5}{4} = \frac{y - 7}{4} = \frac{z + 3}{- 5} and \frac{x - 8}{7} = \frac{y - 4}{1} = \frac{3 - 5}{3}\]


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 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 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). 


A (1, 0, 4), B (0, −11, 3), C (2, −3, 1) are three points and D is the foot of perpendicular from A on BC. Find the coordinates of D


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 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 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 \[\overrightarrow{\beta}\] .


The equations of a line are given by \[\frac{4 - x}{3} = \frac{y + 3}{3} = \frac{z + 2}{6} .\]  Write the direction cosines of a line parallel to this line.


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


The direction ratios of the line x − y + z − 5 = 0 = x − 3y − 6 are proportional to

 

 


Find the value of p for which the following lines are perpendicular : 

`(1-x)/3 = (2y-14)/(2p) = (z-3)/2 ; (1-x)/(3p) = (y-5)/1 = (6-z)/5`


Find the value of λ, so that the lines `(1-"x")/(3) = (7"y" -14)/(λ) = (z -3)/(2) and (7 -7"x")/(3λ) = ("y" - 5)/(1) = (6 -z)/(5)` are at right angles. Also, find whether the lines are intersecting or not.


The equation 4x2 + 4xy + y2 = 0 represents two ______ 


Find the position vector of a point A in space such that `vec"OA"` is inclined at 60º to OX and at 45° to OY and `|vec"OA"|` = 10 units.


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×