हिंदी

By Computing the Shortest Distance Determine Whether the Following Pairs of Lines Intersect Or Not: X − 1 2 = Y + 1 3 = Z a N D X + 1 5 = Y − 2 1 ; Z = 2

Advertisements
Advertisements

प्रश्न

By computing the shortest distance determine whether the following pairs of lines intersect or not: \[\frac{x - 1}{2} = \frac{y + 1}{3} = z \text{ and } \frac{x + 1}{5} = \frac{y - 2}{1}; z = 2\]

योग
Advertisements

उत्तर

\[\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 0}{1} and \frac{x + 1}{5} = \frac{y - 2}{1} = \frac{z - 2}{0}\]

Since the first line passes through the point (1, -1, 0) and has direction ratios proportional to 2, 3, 1, its vector equation is

\[\overrightarrow{r} = \overrightarrow{a_1} + \lambda \overrightarrow{b_1} . . . (1) \]

\[\text{ Here } , \]

\[ \overrightarrow{a_1} = \hat{i}  - \hat{j}  + 0 \hat{k}  \]

\[ \overrightarrow{b_1} = 2 \hat{i}  + 3 \hat{j}  + \hat{k} \]

Also, the second line passes through the point ( -1, 2, 2) and has direction ratios proportional to 5, 1, 0. Its vector equation is

\[\overrightarrow{r} = \overrightarrow{a_2} + \mu \overrightarrow{b_2} . . . (2) \]

\[\text{ Here }, \]

\[ \overrightarrow{a_2} = - \hat{i} + 2 \hat{j} + 2 \hat{k} \]

\[ \overrightarrow{b_2} = 5 \hat{i} + \hat{j}  + 0 \hat{k} \]

Now,

\[\overrightarrow{a_2} - \overrightarrow{a_1} = - 2 \hat{i} + 3 \hat{j}  + 2 \hat{k}  \]

\[\text{ and } \overrightarrow{b_1} \times \overrightarrow{b_2} = \begin{vmatrix}\hat{i}  & \hat{j}  & \hat{k}  \\ 2 & 3 & 1 \\ 5 & 1 & 0\end{vmatrix}\]

\[ = - \hat{i}  + 5 \hat{j} - 13 \hat{k}  \]

\[\left( \overrightarrow{a_2} - \overrightarrow{a_1} \right) . \left( \overrightarrow{b_1} \times \overrightarrow{b_2} \right) = \left( - 2 \hat{i} + 3 \hat{j} + 2 \hat{k} \right) . \left( - \hat{i} + 5 \hat{j} - 13 \hat{k} \right)\]

\[ = 2 + 15 - 26\]

\[ = - 9\]

\[\text{ We observe } \]

\[\left( \overrightarrow{a_2} - \overrightarrow{a_1} \right) . \left( \overrightarrow{b_1} \times \overrightarrow{b_2} \right) \neq 0\]

\[\text{ Thus, the given lines do not intersect } .\]

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 27: Straight Line in Space - Exercise 28.5 [पृष्ठ ३८]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 27 Straight Line in Space
Exercise 28.5 | Q 3.3 | पृष्ठ ३८

वीडियो ट्यूटोरियलVIEW ALL [4]

संबंधित प्रश्न

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


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


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 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. 


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


Find the angle between the following pair of line: 

\[\overrightarrow{r} = \lambda\left( \hat{i} + \hat{j} + 2 \hat{k} \right) \text{ and } \overrightarrow{r} = 2 \hat{j} + \mu\left\{ \left( \sqrt{3} - 1 \right) \hat{i} - \left( \sqrt{3} + 1 \right) \hat{j} + 4 \hat{k} \right\}\]

 


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


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


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 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.  


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. 


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


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


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


Write the vector equation of a line passing through a point having position vector  \[\overrightarrow{\alpha}\] and parallel to vector \[\overrightarrow{\beta}\] .


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

 

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


The lines `x/1 = y/2 = z/3 and (x - 1)/-2 = (y - 2)/-4 = (z - 3)/-6` are


If a line makes angle \[\frac{\pi}{3} \text{ and } \frac{\pi}{4}\]  with x-axis and y-axis respectively, then the angle made by the line with z-axis is


Choose correct alternatives:

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


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


Find the joint equation of pair of lines through the origin which is perpendicular to the lines represented by 5x2 + 2xy - 3y2 = 0 


The equation of line passing through (3, -1, 2) and perpendicular to the lines `overline("r")=(hat"i"+hat"j"-hat"k")+lambda(2hat"i"-2hat"j"+hat"k")` and `overline("r")=(2hat"i"+hat"j"-3hat"k")+mu(hat"i"-2hat"j"+2hat"k")` is ______.


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.


In the vector equation \[\vec{r}=\vec{a}+\lambda\vec{b}\], what does \[\vec{r}\] represent?


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×