English

The lines andx1=y2=z3andx-1-2=y-2-4=z-3-6 are - Mathematics

Advertisements
Advertisements

Question

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

Options

  • coincident 

  • skew

  • intersecting

  • parallel

MCQ

Solution

(a) coincident
The equation of the given lines are 

\[\frac{x}{1} = \frac{y}{2} = \frac{z}{3} . . . (1)\]

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

\[ = \frac{x - 1}{1} = \frac{y - 2}{2} = \frac{z - 3}{3} . . . (2)\]

Thus, the two lines are parallel to the vector  \[\overrightarrow{b} = \hat{i} + 2 \hat{j} + 3 \hat{k} \] and pass through the points (0, 0, 0) and (1, 2, 3).

Now,

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

\[ = \overrightarrow{0} \left[ \because \overrightarrow{a} \times \overrightarrow{a} = \vec{0} \right]\]

Since the distance between the two parallel line is 0, the given lines are coincident.

shaalaa.com
  Is there an error in this question or solution?
Chapter 28: Straight Line in Space - MCQ [Page 42]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 28 Straight Line in Space
MCQ | Q 2 | Page 42

RELATED QUESTIONS

The Cartesian equations of line are 3x+1=6y-2=1-z find its equation in vector form.

 


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


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`


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.


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`.


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


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 in vector form as well as in cartesian form, the equation of the line passing through the points A (1, 2, −1) and B (2, 1, 1).


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.


The cartesian equations of a line are \[\frac{x - 5}{3} = \frac{y + 4}{7} = \frac{z - 6}{2} .\]  Find a vector equation for the line.


Find the points on the line \[\frac{x + 2}{3} = \frac{y + 1}{2} = \frac{z - 3}{2}\]  at a distance of 5 units from the point P (1, 3, 3).


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 through the points (1, −1, 2) and (3, 4, −2) is perpendicular to the through the points (0, 3, 2) and (3, 5, 6).


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 pairs of lines with direction ratios proportional to 5, −12, 13 and −3, 4, 5


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


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


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.


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 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 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 vector equations are: \[\overrightarrow{r} = \left( 8 + 3\lambda \right) \hat{i} - \left( 9 + 16\lambda \right) \hat{j} + \left( 10 + 7\lambda \right) \hat{k} \]\[\overrightarrow{r} = 15 \hat{i} + 29 \hat{j} + 5 \hat{k} + \mu\left( 3 \hat{i}  + 8 \hat{j} - 5 \hat{k}  \right)\]


Write the vector equations of the following lines and hence determine the distance between them  \[\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z + 4}{6} \text{ and } \frac{x - 3}{4} = \frac{y - 3}{6} = \frac{z + 5}{12}\]


Find the shortest distance between the lines \[\overrightarrow{r} = \hat{i} + 2 \hat{j} + 3 \hat{k} + \lambda\left( \hat{i} - 3 \hat{j} + 2 \hat{k} \right) \text{ and }  \overrightarrow{r} = 4 \hat{i} + 5 \hat{j}  + 6 \hat{k} + \mu\left( 2 \hat{i} + 3 \hat{j} + \hat{k} \right)\]


Write the value of λ for which the lines  \[\frac{x - 3}{- 3} = \frac{y + 2}{2\lambda} = \frac{z + 4}{2} \text{ and } \frac{x + 1}{3\lambda} = \frac{y - 2}{1} = \frac{z + 6}{- 5}\]  are perpendicular to each other.


Write the formula for the shortest distance between the lines 

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

 


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 equation of a plane which passes through the point (3, 2, 0) and contains the line \[\frac{x - 3}{1} = \frac{y - 6}{5} = \frac{z - 4}{4}\].

 

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`


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


If 2x + y = 0 is one of the line represented by 3x2 + kxy + 2y2 = 0 then k = ______ 


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`


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×