हिंदी

By-computing-shortest-distance-determine-whether-following-pairs-lines-intersect-or-not-r-i-j-k-3-i-j-r-4-i-k-2-i-3-k

Advertisements
Advertisements

प्रश्न

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

योग
Advertisements

उत्तर १

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

Comparing the given equations with the equations 

\[\overrightarrow{r} = \overrightarrow{a_1} + \lambda \overrightarrow{b_1} \text{ and }  \overrightarrow{r} = \overrightarrow{a_2} + \mu \overrightarrow{b_2}\]

We get , 

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

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

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

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

\[ \therefore \overrightarrow{a_2} - \overrightarrow{a_1} = \hat{i} \]

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

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

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

\[ = - 1\]

\[\text{ We observe } \]

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

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

shaalaa.com

उत्तर २

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

Comparing the given equations with the equations 

\[\vec{r} = \vec{a_1} + \lambda \vec{b_1} \text{ and }  \vec{r} = \vec{a_2} + \mu \vec{b_2}\]

\[\vec{a_1} = \hat{i} - \hat{j} \]

\[ \vec{a_2} = 2 \hat{i} - \hat{j} \]

\[ \vec{b_1} = 2 \hat{i} + \hat{k} \]

\[ \vec{b_2} = \hat{i} + \hat{j} - \hat{k} \]

\[ \therefore \vec{a_2} - \vec{a_1} = \hat{i} \]

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

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

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

\[ = - 1\]

\[\text{ We observe } \]

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

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

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 3.1 | पृष्ठ ३८

वीडियो ट्यूटोरियलVIEW ALL [4]

संबंधित प्रश्न

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


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 (2, −1, 1) and parallel to the line whose equations are \[\frac{x - 3}{2} = \frac{y + 1}{7} = \frac{z - 2}{- 3} .\]


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


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


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: 

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

 


Find the angle between the following pair of line:

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


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 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 angle between the pairs of lines with direction ratios proportional to  2, 2, 1 and 4, 1, 8 .

 


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.  


Determine whether the following pair of lines intersect or not: 

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


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.


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


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 coordinates of the foot of perpendicular drawn from the point A(1, 8, 4) to the line joining the points B(0, −1, 3) and C(2, −3, −1).      


Find the shortest distance between the following pairs of lines whose vector equations are: \[\vec{r} = 3 \hat{i} + 8 \hat{j} + 3 \hat{k}  + \lambda\left( 3 \hat{i}  - \hat{j}  + \hat{k}  \right) \text{ and }  \vec{r} = - 3 \hat{i}  - 7 \hat{j}  + 6 \hat{k}  + \mu\left( - 3 \hat{i}  + 2 \hat{j}  + 4 \hat{k} \right)\]


By computing the shortest distance determine whether the following pairs of lines intersect or not: \[\frac{x - 5}{4} = \frac{y - 7}{- 5} = \frac{z + 3}{- 5} \text{ and } \frac{x - 8}{7} = \frac{y - 7}{1} = \frac{z - 5}{3}\]


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


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


Cartesian equations of a line AB are  \[\frac{2x - 1}{2} = \frac{4 - y}{7} = \frac{z + 1}{2} .\]   Write the direction ratios of a line parallel to AB.


Find the angle between the lines 2x=3y=-z and 6x =-y=-4z.

 


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 

 


If the direction ratios of a line are proportional to 1, −3, 2, then its direction cosines 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


 The equation of a line is 2x -2 = 3y +1 = 6z -2 find the direction ratios and also find the vector equation of the line. 


If y – 2x – k = 0 touches the conic 3x2 – 5y2 = 15, find the value of k. 


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


Auxillary equation of 2x2 + 3xy − 9y2 = 0 is ______ 


Equation of a line passing through (1, 1, 1) and parallel to z-axis is ______.


A line passes through the point (2, – 1, 3) and is perpendicular to the lines `vecr = (hati + hatj - hatk) + λ(2hati - 2hatj + hatk)` and `vecr = (2hati - hatj - 3hatk) + μ(hati + 2hatj + 2hatk)` obtain its equation.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×