मराठी

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`.

बेरीज
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
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 11: Three Dimensional Geometry - Exercise 11.1 [पृष्ठ ४७७]

APPEARS IN

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

व्हिडिओ ट्यूटोरियलVIEW ALL [4]

संबंधित प्रश्‍न

Find the separate equations of the lines represented by the equation 3x2 – 10xy – 8y2 = 0.


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 the Cartesian equations of the lines that pass through the origin and (5, −2, 3).


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


ABCD is a parallelogram. The position vectors of the points AB and C are respectively, \[4 \hat{ i} + 5 \hat{j} -10 \hat{k} , 2 \hat{i} - 3 \hat{j} + 4 \hat{k}  \text{ and } - \hat{i} + 2 \hat{j} + \hat{k} .\]  Find the vector equation of the line BD. Also, reduce it to cartesian form.


Find the vector equation of a line passing through the point with position vector  \[\hat{i} - 2 \hat{j} - 3 \hat{k}\]  and parallel to the line joining the points with position vectors  \[\hat{i} - \hat{j} + 4 \hat{k} \text{ and } 2 \hat{i} + \hat{j} + 2 \hat{k} .\] Also, find the cartesian equivalent of this equation.


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.


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 joining the origin to the point (2, 1, 1) is perpendicular to the line determined by the points (3, 5, −1) and (4, 3, −1). 


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:

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


Find the equation of the line passing through the point (1, 2, −4) and parallel to the line \[\frac{x - 3}{4} = \frac{y - 5}{2} = \frac{z + 1}{3} .\] 


Find the equations of the line passing through the point (−1, 2, 1) and parallel to the line  \[\frac{2x - 1}{4} = \frac{3y + 5}{2} = \frac{2 - z}{3} .\]


Determine the equations of the line passing through the point (1, 2, −4) and perpendicular to the two lines \[\frac{x - 8}{8} = \frac{y + 9}{- 16} = \frac{z - 10}{7} \text{    and    } \frac{x - 15}{3} = \frac{y - 29}{8} = \frac{z - 5}{- 5}\]


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 value of λ so that the following lines are perpendicular to each other. \[\frac{x - 5}{5\lambda + 2} = \frac{2 - y}{5} = \frac{1 - z}{- 1}, \frac{x}{1} = \frac{2y + 1}{4\lambda} = \frac{1 - z}{- 3}\]


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.  


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. 


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 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 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: \[\vec{r} = 3 \hat{i} + 8 \hat{j} + 3 \hat{k}  + \lambda\left( 3 \hat{i}  - \hat{j}  + \hat{k}  \right) \text{ and }  \vec{r} = - 3 \hat{i}  - 7 \hat{j}  + 6 \hat{k}  + \mu\left( - 3 \hat{i}  + 2 \hat{j}  + 4 \hat{k} \right)\]


Find the shortest distance between the following pairs of lines whose vector equations are: \[\overrightarrow{r} = \left( \hat{i} + 2 \hat{j} + 3 \hat{k} \right) + \lambda\left( 2 \hat{i} + 3 \hat{j}  + 4 \hat{k}  \right) \text{ and }  \overrightarrow{r} = \left( 2 \hat{i} + 4 \hat{j} + 5 \hat{k} \right) + \mu\left( 3 \hat{i}  + 4 \hat{j}  + 5 \hat{k} \right)\]


Find the shortest distance between the following pairs of lines whose vector equations are: \[\overrightarrow{r} = \left( 1 - t \right) \hat{i} + \left( t - 2 \right) \hat{j} + \left( 3 - t \right) \hat{k}  \text{ and }  \overrightarrow{r} = \left( s + 1 \right) \hat{i}  + \left( 2s - 1 \right) \hat{j}  - \left( 2s + 1 \right) \hat{k} \]


Write the cartesian and vector equations of Y-axis.

 

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


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 

\[\vec{r} = \left( 2 \hat{i}  - 5 \hat{j}  + \hat{k}  \right) + \lambda\left( 3 \hat{i}  + 2 \hat{j}  + 6 \hat{k}  \right)\] and \[\vec{r} = 7 \hat{i} - 6 \hat{k}  + \mu\left( \hat{i}  + 2 \hat{j}  + 2 \hat{k}  \right)\] 


The 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 perpendicular distance of the point P (1, 2, 3) from the line \[\frac{x - 6}{3} = \frac{y - 7}{2} = \frac{z - 7}{- 2}\] is 

 


The equation of the line passing through the points \[a_1 \hat{i}  + a_2 \hat{j}  + a_3 \hat{k}  \text{ and }  b_1 \hat{i} + b_2 \hat{j}  + b_3 \hat{k} \]  is 


If a line makes angles α, β and γ with the axes respectively, then cos 2 α + cos 2 β + cos 2 γ =


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 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.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×