English

The Cartesian Equations of a Line Are X − 5 3 = Y + 4 7 = Z − 6 2 . Find a Vector Equation for the Line. - Mathematics

Advertisements
Advertisements

Question

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.

Sum
Advertisements

Solution

The cartesian equation of the given line is \[\frac{x - 5}{3} = \frac{y + 4}{7} = \frac{z - 6}{2}\] 

It can be re-written as 

\[\frac{x - 5}{3} = \frac{y - \left( - 4 \right)}{7} = \frac{z - 6}{2}\] 

Thus, the given line passes through the point having position vector   \[\vec{a} = 5 \hat{i} - 4 \hat{j} + 6 \hat{k}\] and is parallel to the vector  \[\vec{b} = 3 \hat{i}  + 7 \hat{j}+ 2 \hat{k} \] 

We know that the vector equation of a line passing through a point with position vector  `vec a ` and parallel to the vector `vec b` is  \[\vec{r} = \vec{a} + \lambda \vec{b}\] 

Vector equation of the required line is 

\[\vec{r} = \left( 5 \hat{i} - 4 \hat{j} + 6 \hat{k} \right) + \lambda \left( 3 \hat{i} + 7 \hat{j} + 2 \hat{k} \right)\]

\[\text{ Here} , \lambda \text{ is a parameter } . \]

shaalaa.com
  Is there an error in this question or solution?
Chapter 28: Straight Line in Space - Exercise 28.1 [Page 10]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 28 Straight Line in Space
Exercise 28.1 | Q 9 | Page 10

RELATED QUESTIONS

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


Find the coordinates of the point where the line through the points A(3, 4, 1) and B(5, 1, 6) crosses the XZ plane. Also find the angle which this line makes with the XZ plane.


If the Cartesian equations of a line are ` (3-x)/5=(y+4)/7=(2z-6)/4` , write the vector equation for the line.


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 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:

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


If the lines \[\frac{x - 1}{- 3} = \frac{y - 2}{2 \lambda} = \frac{z - 3}{2} \text{     and     } \frac{x - 1}{3\lambda} = \frac{y - 1}{1} = \frac{z - 6}{- 5}\]  are perpendicular, find the value of λ.


If the coordinates of the points ABCD be (1, 2, 3), (4, 5, 7), (−4, 3, −6) and (2, 9, 2) respectively, then find the angle between the lines AB and CD


Show that the lines \[\frac{x - 1}{3} = \frac{y + 1}{2} = \frac{z - 1}{5} \text{           and                } \frac{x + 2}{4} = \frac{y - 1}{3} = \frac{z + 1}{- 2}\]  do not intersect. 


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.


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 perpendicular distance of the point (1, 0, 0) from the line  \[\frac{x - 1}{2} = \frac{y + 1}{- 3} = \frac{z + 10}{8}.\] Also, find the coordinates of the foot of the perpendicular and the equation of the perpendicular.


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


Find the equation of the perpendicular drawn from the point P (−1, 3, 2) to the line  \[\overrightarrow{r} = \left( 2 \hat{j} + 3 \hat{k} \right) + \lambda\left( 2 \hat{i} + \hat{j} + 3 \hat{k}  \right) .\]  Also, find the coordinates of the foot of the perpendicular from P.


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


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


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


Write the coordinate axis to which the line \[\frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 1}{0}\]  is  perpendicular.


Find the angle between the lines 

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


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 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


The direction ratios of the line x − y + z − 5 = 0 = x − 3y − 6 are proportional to

 

 


The equation of the line passing through the points \[a_1 \hat{i}  + a_2 \hat{j}  + a_3 \hat{k}  \text{ and }  b_1 \hat{i} + b_2 \hat{j}  + b_3 \hat{k} \]  is 


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 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`


Choose correct alternatives:

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


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 ______.


The distance of the point (4, 3, 8) from the Y-axis is ______.


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`


P is a point on the line joining the points A(0, 5, −2) and B(3, −1, 2). If the x-coordinate of P is 6, then its z-coordinate is ______.


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`.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×