मराठी

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

प्रश्न

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.

बेरीज
Advertisements

उत्तर

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
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 28: Straight Line in Space - Exercise 28.1 [पृष्ठ १०]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 28 Straight Line in Space
Exercise 28.1 | Q 9 | पृष्ठ १०

व्हिडिओ ट्यूटोरियलVIEW ALL [4]

संबंधित प्रश्‍न

Find the vector and Cartesian equations of the line through the point (1, 2, −4) and perpendicular to the two lines. 

`vecr=(8hati-19hatj+10hatk)+lambda(3hati-16hatj+7hatk) " and "vecr=(15hati+29hatj+5hatk)+mu(3hati+8hatj-5hatk)`

 

 


 

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 the Cartesian equations of the lines that pass through the origin and (5, −2, 3).


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


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


The cartesian equations of a line are x = ay + bz = cy + d. Find its direction ratios and reduce it to vector form. 


Find the vector equation of a line passing through the point with position vector  \[\hat{i} - 2 \hat{j} - 3 \hat{k}\]  and parallel to the line joining the points with position vectors  \[\hat{i} - \hat{j} + 4 \hat{k} \text{ and } 2 \hat{i} + \hat{j} + 2 \hat{k} .\] Also, find the cartesian equivalent of this equation.


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


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). 


Determine whether the following pair of lines intersect or not: 

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


Determine whether the following pair of lines intersect or not: 

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


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 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: \[\overrightarrow{r} = \left( 1 - t \right) \hat{i} + \left( t - 2 \right) \hat{j} + \left( 3 - t \right) \hat{k}  \text{ and }  \overrightarrow{r} = \left( s + 1 \right) \hat{i}  + \left( 2s - 1 \right) \hat{j}  - \left( 2s + 1 \right) \hat{k} \]


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 lines whose cartesian equations are : \[\frac{x - 1}{2} = \frac{y + 1}{3} = z \text{ and } \frac{x + 1}{3} = \frac{y - 2}{1}; z = 2\]


Find the shortest distance between the following pairs of lines whose cartesian equations are : \[\frac{x - 1}{- 1} = \frac{y + 2}{1} = \frac{z - 3}{- 2} \text{ and } \frac{x - 1}{1} = \frac{y + 1}{2} = \frac{z + 1}{- 2}\]


Find the shortest distance between the following pairs of lines whose cartesian equations are:  \[\frac{x - 3}{1} = \frac{y - 5}{- 2} = \frac{z - 7}{1} \text{ and } \frac{x + 1}{7} = \frac{y + 1}{- 6} = \frac{z + 1}{1}\]


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


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 cartesian and vector equations of X-axis.

 

Write the cartesian and vector equations of Z-axis.

 

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 \[\frac{x - 2}{2} = \frac{2y - 5}{- 3}, z = 2 .\]


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


The cartesian equations of a line AB are  \[\frac{2x - 1}{\sqrt{3}} = \frac{y + 2}{2} = \frac{z - 3}{3} .\]   Find the direction cosines of a line parallel to AB


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

 


 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. 


The separate equations of the lines represented by `3x^2 - 2sqrt(3)xy - 3y^2` = 0 are ______ 


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


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 ______.


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×