हिंदी

Find the Cartesian Equation of a Line Passing Through (1, −1, 2) and Parallel to the Line Whose Equations Are X − 3 1 = Y − 1 2 = Z + 1 − 2 Also, Reduce the Equation Obtained in Vector Form.

Advertisements
Advertisements

प्रश्न

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.

योग
Advertisements

उत्तर

We know that the cartesian equation of a line passing through a point with position vector `vec a `  and parallel to the vector `vec m ` is  \[\frac{x - x_1}{a} = \frac{y - y_2}{b} = \frac{z - z_3}{c}\] 

Here, 

\[\vec{a} = x_1 \hat{i} + y_1 \hat{j} + z_1 \hat{k} \]

\[ \vec{m} = a \hat{i} + b \hat{j} + c \hat{k} \]

Here,  \[\vec{a} = \hat{i} - \hat{j} + 2 \hat{k} \text{ and } \vec{b} = \hat{i} + 2 \hat{j} - 2 \hat{k}\] 

Cartesian equation of the required line is 

\[\frac{x - 1}{1} = \frac{y - \left( - 1 \right)}{2} = \frac{z - 2}{- 2}\]

\[ \Rightarrow \frac{x - 1}{1} = \frac{y + 1}{2} = \frac{z - 2}{- 2}\]

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

Here, the line is passing through the point  (1,1,-2)  and its direction ratios are proportional to 1, 2, -2 .

Vector equation of the required line is

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

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 27: Straight Line in Space - Exercise 28.1 [पृष्ठ १०]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 27 Straight Line in Space
Exercise 28.1 | Q 10 | पृष्ठ १०

वीडियो ट्यूटोरियलVIEW ALL [4]

संबंधित प्रश्न

Find the separate equations of the lines represented by the equation 3x2 – 10xy – 8y2 = 0.


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


 

A line passes through (2, −1, 3) and is perpendicular to the lines `vecr=(hati+hatj-hatk)+lambda(2hati-2hatj+hatk) and vecr=(2hati-hatj-3hatk)+mu(hati+2hatj+2hatk)` . Obtain its equation in vector and Cartesian from. 

 

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`.


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


Find the vector and the Cartesian equations of the line that passes through the points (3, −2, −5), (3, −2, 6).

 


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


Show that the line joining the origin to the point (2, 1, 1) is perpendicular to the line determined by the points (3, 5, – 1), (4, 3, – 1).


Find the vector equation of the lines which passes through the point with position vector `4hati - hatj +2hatk` and is in the direction of `-2hati + hatj + hatk`


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


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


Show that the points whose position vectors are  \[- 2 \hat{i} + 3 \hat{j} , \hat{i} + 2 \hat{j} + 3 \hat{k}  \text{ and }  7 \text{ i}  - \text{ k} \]  are collinear.


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


Show that the line joining the origin to the point (2, 1, 1) is perpendicular to the line determined by the points (3, 5, −1) and (4, 3, −1). 


Find the angle between the following pair of line: 

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


Find the equation of the line passing through the point (2, −1, 3) and parallel to the line  \[\overrightarrow{r} = \left( \hat{i} - 2 \hat{j} + \hat{k} \right) + \lambda\left( 2 \hat{i} + 3 \hat{j} - 5 \hat{k} \right) .\]


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 vector equation of the line passing through the point (2, −1, −1) which is parallel to the line 6x − 2 = 3y + 1 = 2z − 2. 


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


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.


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:  

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


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


Find the shortest distance between the following pairs of lines whose vector equations are: \[\overrightarrow{r} = \left( 8 + 3\lambda \right) \hat{i} - \left( 9 + 16\lambda \right) \hat{j} + \left( 10 + 7\lambda \right) \hat{k} \]\[\overrightarrow{r} = 15 \hat{i} + 29 \hat{j} + 5 \hat{k} + \mu\left( 3 \hat{i}  + 8 \hat{j} - 5 \hat{k}  \right)\]


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

 

 


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 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


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

 


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


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


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

 

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


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


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×