English

If the Direction Ratios of a Line Are Proportional to 1, −3, 2, Then Its Direction Cosines Are (A) 1 √ 14 , − 3 √ 14 , 2 √ 14 (B) 1 √ 14 , 2 √ 14 , 3 √ 14 (C) − 1 √ 14 , 3 √ 14 , 2 √ 14

Advertisements
Advertisements

Question

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

 

Options

  •  \[\frac{1}{\sqrt{14}}, - \frac{3}{\sqrt{14}}, \frac{2}{\sqrt{14}}\] 

  •  \[\frac{1}{\sqrt{14}}, \frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}}\] 

  •  \[- \frac{1}{\sqrt{14}}, \frac{3}{\sqrt{14}}, \frac{2}{\sqrt{14}}\] 

  •  \[- \frac{1}{\sqrt{14}}, - \frac{2}{\sqrt{14}}, - \frac{3}{\sqrt{14}}\]

MCQ
Advertisements

Solution

 \[\frac{1}{\sqrt{14}}, - \frac{3}{\sqrt{14}}, \frac{2}{\sqrt{14}}\] 

The direction ratios of the line are proportional to 1, -3, 2 .

∴  The direction cosines of the line are

\[\frac{1}{\sqrt{1^2 + \left( - 3 \right)^2 + 2^2}}, \frac{- 3}{\sqrt{1^2 + \left( - 3 \right)^2 + 2^2}}, \frac{2}{\sqrt{1^2 + \left( - 3 \right)^2 + 2^2}} \]

\[ = \frac{1}{\sqrt{14}}, \frac{- 3}{\sqrt{14}}, \frac{2}{\sqrt{14}}\]

shaalaa.com
  Is there an error in this question or solution?
Chapter 27: Straight Line in Space - MCQ [Page 43]

APPEARS IN

R.D. Sharma Mathematics Volume 1 and 2 [English] Class 12
Chapter 27 Straight Line in Space
MCQ | Q 9 | Page 43

RELATED QUESTIONS

 

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.

 

Find the vector and Cartesian equations of a line passing through (1, 2, –4) and perpendicular to the two lines `(x - 8)/3 = (y + 19)/(-16) = (z - 10)/7` and `(x - 15)/3 = (y - 29)/8 = (z - 5)/(-5)`


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


A line passes through the point with position vector \[2 \hat{i} - 3 \hat{j} + 4 \hat{k} \] and is in the direction of  \[3 \hat{i} + 4 \hat{j} - 5 \hat{k} .\] Find equations of the line in vector and cartesian form. 


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.


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


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 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).


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


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 pairs of lines with direction ratios proportional to  1, 2, −2 and −2, 2, 1 .


Find the angle between two lines, one of which has direction ratios 2, 2, 1 while the  other one is obtained by joining the points (3, 1, 4) and (7, 2, 12). 


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.


Determine whether the following pair of lines intersect or not:  

\[\frac{x - 5}{4} = \frac{y - 7}{4} = \frac{z + 3}{- 5} and \frac{x - 8}{7} = \frac{y - 4}{1} = \frac{3 - 5}{3}\]


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 length of the perpendicular drawn from the point (5, 4, −1) to the line \[\overrightarrow{r} = \hat{i}  + \lambda\left( 2 \hat{i} + 9 \hat{j} + 5 \hat{k} \right) .\]


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


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


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


Find the distance between the lines l1 and l2 given by  \[\overrightarrow{r} = \hat{i} + 2 \hat{j} - 4 \hat{k} + \lambda\left( 2 \hat{i}  + 3 \hat{j}  + 6 \hat{k}  \right) \text{ and } , \overrightarrow{r} = 3 \hat{i} + 3 \hat{j}  - 5 \hat{k}  + \mu\left( 2 \hat{i} + 3 \hat{j}  + 6 \hat{k}  \right)\]

 

 


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.


Write the direction cosines of the line \[\frac{x - 2}{2} = \frac{2y - 5}{- 3}, z = 2 .\]


Write the angle between the lines \[\frac{x - 5}{7} = \frac{y + 2}{- 5} = \frac{z - 2}{1} \text{ and } \frac{x - 1}{1} = \frac{y}{2} = \frac{z - 1}{3} .\]


Write the vector equation of a line given by \[\frac{x - 5}{3} = \frac{y + 4}{7} = \frac{z - 6}{2} .\]

 


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 projections of a line segment on XY and Z axes are 12, 4 and 3 respectively. The length and direction cosines of the line segment are


The shortest distance between the lines  \[\frac{x - 3}{3} = \frac{y - 8}{- 1} = \frac{z - 3}{1} \text{ and }, \frac{x + 3}{- 3} = \frac{y + 7}{2} = \frac{z - 6}{4}\] 

 

 

 

 


Find the value of  λ for which the following lines are perpendicular to each other: 

`(x - 5)/(5 lambda + 2 ) = ( 2 - y )/5 = (1 - z ) /-1 ; x /1 = ( y + 1/2)/(2 lambda ) = ( z -1 ) / 3`


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


Choose correct alternatives:

The difference between the slopes of the lines represented by 3x2 - 4xy + y2 = 0 is 2


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`.


Find the vector equation of the lines passing through the point having position vector `(-hati - hatj + 2hatk)` and parallel to the line `vecr = (hati + 2hatj + 3hatk) + λ(3hati + 2hatj + hatk)`.


Find the equations of the diagonals of the parallelogram PQRS whose vertices are P(4, 2, – 6), Q(5, – 3, 1), R(12, 4, 5) and S(11, 9, – 2). Use these equations to find the point of intersection of diagonals.


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


Which Cartesian equation is obtained for the line through \[(5,2,-4)\] parallel to \[3\hat{i}+2\hat{j}-8\hat{k}\]?


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×