मराठी

Show that the Lines X + 1 3 = Y + 3 5 = Z + 5 7 a N D X − 2 1 = Y − 4 3 = Z − 6 5 Intersect. Find Their Point of Intersection.

Advertisements
Advertisements

प्रश्न

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.

बेरीज
Advertisements

उत्तर

The coordinates of any point on the first line are given by 

\[\frac{x + 1}{3} = \frac{y + 3}{5} = \frac{z + 5}{7} = \lambda\]

\[ \Rightarrow x = 3\lambda - 1\]

\[ y = 5\lambda - 3\]

\[ z = 7\lambda - 5\] 

The coordinates of a general point on the first line are 

\[\left( 3\lambda - 1, 5\lambda - 3, 7\lambda - 5 \right)\] 

The coordinates of any point on the second line are given by 

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

\[ \Rightarrow x = \mu + 2\]

\[ y = 3\mu + 4 \]

\[ z = 5\mu + 6\]

The coordinates of a general point on the second line are

\[\left( \mu + 2, 3\mu + 4, 5\mu + 6 \right)\] 

If the lines intersect, then they have a common point. So, for some values of  \[\lambda \text{ and } \mu\] we must have ,

\[3\lambda - 1 = \mu + 2, 5\lambda - 3 = 3\mu + 4, 7\lambda - 5 = 5\mu + 6\]

\[ \Rightarrow 3\lambda - \mu = 3 . . . (1)\]

\[ 5\lambda - 3\mu = 7 . . . (2)\]

\[ 7\lambda - 5\mu = 11 . . . (3)\]

\[\text{ Solving (1) and (2), we get } \]

\[\lambda = \frac{1}{2} \]

\[\mu = - \frac{3}{2}\]

\[\text { Substituting }  \lambda = \frac{1}{2} \text{ and } \mu = - \frac{3}{2} \text { in (3), we get } \]

\[LHS = 7\lambda - 5\mu\]

\[ = 7\left( \frac{1}{2} \right) - 5\left( - \frac{3}{2} \right)\]

\[ = 11\]

\[ = RHS\]

\[\text{ Since } \lambda = \frac{1}{2} \text{ and } \mu = - \frac{3}{2} \text{ satisfy (3), the given lines intersect .}  \]

\[\text{ Substituting the value of       } \lambda \text{ in the general coordinates of the first line, we get } \]

\[x = \frac{1}{2}\]

\[y = - \frac{1}{2}\]

\[z = - \frac{3}{2}\]

\[\text{ Hence, the given lines intersect at point }  \left( \frac{1}{2}, - \frac{1}{2}, - \frac{3}{2} \right) .\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 27: Straight Line in Space - Exercise 28.3 [पृष्ठ २२]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
पाठ 27 Straight Line in Space
Exercise 28.3 | Q 3 | पृष्ठ २२

व्हिडिओ 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.


Find the coordinates of the point where the line through the points A(3, 4, 1) and B(5, 1, 6) crosses the XZ plane. Also find the angle which this line makes with the XZ plane.


 

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.


Show that the line through the points (4, 7, 8) (2, 3, 4) is parallel to the line through the points (−1, −2, 1), (1, 2, 5).


Show that the lines `(x-5)/7 = (y + 2)/(-5) = z/1` and `x/1 = y/2 = z/3` are perpendicular to each other.


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


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 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.


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.


Find the vector equation of the line passing through the point A(1, 2, –1) and parallel to the line 5x – 25 = 14 – 7y = 35z.


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: 

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


Find the angle between the following pair of line:

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


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


Find the equation of the line passing through the point  \[\hat{i}  + \hat{j}  - 3 \hat{k} \] and perpendicular to the lines  \[\overrightarrow{r} = \hat{i}  + \lambda\left( 2 \hat{i} + \hat{j}  - 3 \hat{k}  \right) \text { and }  \overrightarrow{r} = \left( 2 \hat{i}  + \hat{j}  - \hat{ k}  \right) + \mu\left( \hat{i}  + \hat{j}  + \hat{k}  \right) .\]

  

 

 

 


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 \[\vec{r} = 3 \hat{i} + 2 \hat{j} - 4 \hat{k} + \lambda\left( \hat{i} + 2 \hat{j} + 2 \hat{k} \right) \text{ and } \vec{r} = 5 \hat{i} - 2 \hat{j}  + \mu\left( 3 \hat{i} + 2 \hat{j} + 6 \hat{k} \right)\] are intersecting. Hence, find their point of intersection.


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 the perpendicular from (0, 2, 7) on the line \[\frac{x + 2}{- 1} = \frac{y - 1}{3} = \frac{z - 3}{- 2} .\]


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 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: \[\overrightarrow{r} = \left( 1 - t \right) \hat{i} + \left( t - 2 \right) \hat{j} + \left( 3 - t \right) \hat{k}  \text{ and }  \overrightarrow{r} = \left( s + 1 \right) \hat{i}  + \left( 2s - 1 \right) \hat{j}  - \left( 2s + 1 \right) \hat{k} \]


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


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


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 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) 


Write the direction cosines of the line whose cartesian equations are 6x − 2 = 3y + 1 = 2z − 4.

 

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


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


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 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 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.


Choose correct alternatives:

If the equation 4x2 + hxy + y2 = 0 represents two coincident lines, then h = _______


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×