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. - Mathematics

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 28: Straight Line in Space - Exercise 28.3 [पृष्ठ २२]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 28 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.


The Cartesian equations of line are 3x -1 = 6y + 2 = 1 - z. Find the vector equation of line.


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


 

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.

 

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).


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


Find the cartesian equation of a line passing through (1, −1, 2) and parallel to the line whose equations are  \[\frac{x - 3}{1} = \frac{y - 1}{2} = \frac{z + 1}{- 2}\]  Also, reduce the equation obtained in vector form.


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 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.


Find the cartesian and vector equations of a line which passes through the point (1, 2, 3) and is parallel to the line  \[\frac{- x - 2}{1} = \frac{y + 3}{7} = \frac{2z - 6}{3} .\] 


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). 


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.


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 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 equation of line passing through the points A (0, 6, −9) and B (−3, −6, 3). If D is the foot of perpendicular drawn from a point C (7, 4, −1) on the line AB, then find the coordinates of the point D and the equation of line CD


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 are: \[\overrightarrow{r} = \left( \hat{i} + \hat{j} \right) + \lambda\left( 2 \hat{i} - \hat{j} + \hat{k} \right) \text{ and } , \overrightarrow{r} = 2 \hat{i} + \hat{j} - \hat{k} + \mu\left( 3 \hat{i} - 5 \hat{j} + 2 \hat{k} \right)\]


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:  \[\frac{x - 3}{1} = \frac{y - 5}{- 2} = \frac{z - 7}{1} \text{ and } \frac{x + 1}{7} = \frac{y + 1}{- 6} = \frac{z + 1}{1}\]


By computing the shortest distance determine whether the following pairs of lines intersect or not: \[\overrightarrow{r} = \left( \hat{i} + \hat{j} - \hat{k} \right) + \lambda\left( 3 \hat{i} - \hat{j} \right) \text{ and } \overrightarrow{r} = \left( 4 \hat{i} - \hat{k}  \right) + \mu\left( 2 \hat{i}  + 3 \hat{k} \right)\] 


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


Write the cartesian and vector equations of X-axis.

 

Write the cartesian and vector equations of Y-axis.

 

Write the cartesian and vector equations of Z-axis.

 

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

 

Write the formula for the shortest distance between the lines 

\[\overrightarrow{r} = \overrightarrow{a_1} + \lambda \overrightarrow{b} \text{ and }  \overrightarrow{r} = \overrightarrow{a_2} + \mu \overrightarrow{b} .\] 

 


Write the condition for the lines  \[\vec{r} = \overrightarrow{a_1} + \lambda \overrightarrow{b_1} \text{ and  } \overrightarrow{r} = \overrightarrow{a_2} + \mu \overrightarrow{b_2}\] to be intersecting.


If the equations of a line AB are 

\[\frac{3 - x}{1} = \frac{y + 2}{- 2} = \frac{z - 5}{4},\] write the direction ratios of a line parallel to AB


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

 

If the lines represented by kx2 − 3xy + 6y2 = 0 are perpendicular to each other, then


Find the vector equation of a line passing through a point with position vector `2hati - hatj + hatk` and parallel to the line joining the points `-hati + 4hatj + hatk` and `-hati + 2hatj + 2hatk`.


The lines `(x - 1)/2 = (y + 1)/2 = (z - 1)/4` and `(x - 3)/1 = (y - k)/2 = z/1` intersect each other at point


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×