हिंदी

Write the Vector Equation of a Line Passing Through a Point Having Position Vector → α and Parallel to Vector → β . - Mathematics

Advertisements
Advertisements

प्रश्न

Write the vector equation of a line passing through a point having position vector  \[\overrightarrow{\alpha}\] and parallel to vector \[\overrightarrow{\beta}\] .

टिप्पणी लिखिए

उत्तर

The vector equation of the line passing through the point having position vector  \[\overrightarrow{\alpha}\] and parallel to vector \[\overrightarrow{\beta}\] is \[\overrightarrow{r} = \overrightarrow{\alpha} + \lambda \overrightarrow{\beta}\]

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 28: Straight Line in Space - Very Short Answers [पृष्ठ ४१]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
अध्याय 28 Straight Line in Space
Very Short Answers | Q 4 | पृष्ठ ४१

वीडियो ट्यूटोरियलVIEW ALL [4]

संबंधित प्रश्न

The Cartesian equations of line are 3x -1 = 6y + 2 = 1 - z. Find the vector equation of line.


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

 


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


 

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 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.

 

Show that the lines `(x-5)/7 = (y + 2)/(-5) = z/1` and `x/1 = y/2 = z/3` are perpendicular to each other.


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 which is parallel to the vector \[2 \hat{i} - \hat{j} + 3 \hat{k}\]  and which passes through the point (5, −2, 4). Also, reduce it to cartesian form.


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.


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.


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 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.


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 (4, 7, 8) and (2, 3, 4) is parallel to the line through the points (−1, −2, 1) and, (1, 2, 5).


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: 

\[\overrightarrow{r} = \lambda\left( \hat{i} + \hat{j} + 2 \hat{k} \right) \text{ and } \overrightarrow{r} = 2 \hat{j} + \mu\left\{ \left( \sqrt{3} - 1 \right) \hat{i} - \left( \sqrt{3} + 1 \right) \hat{j} + 4 \hat{k} \right\}\]

 


Find the angle between the pairs of lines with direction ratios proportional to   abc and b − cc − aa − b.


Find the equations of the line passing through the point (−1, 2, 1) and parallel to the line  \[\frac{2x - 1}{4} = \frac{3y + 5}{2} = \frac{2 - z}{3} .\]


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


Determine whether the following pair of lines intersect or not: 

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


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 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 from (0, 2, 7) on the line \[\frac{x + 2}{- 1} = \frac{y - 1}{3} = \frac{z - 3}{- 2} .\]


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


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


Write the angle between the lines 2x = 3y = −z and 6x = −y = −4z.

 

Find the angle between the lines 2x=3y=-z and 6x =-y=-4z.

 


The angle between the lines

\[\frac{x - 1}{1} = \frac{y - 1}{1} = \frac{z - 1}{2} \text{ and }, \frac{x - 1}{- \sqrt{3} - 1} = \frac{y - 1}{\sqrt{3} - 1} = \frac{z - 1}{4}\] is 

If a line makes angle \[\frac{\pi}{3} \text{ and } \frac{\pi}{4}\]  with x-axis and y-axis respectively, then the angle made by the line with z-axis is


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

 

 

 

 


Show that the lines \[\frac{5 - x}{- 4} = \frac{y - 7}{4} = \frac{z + 3}{- 5} \text { and } \frac{x - 8}{7} = \frac{2y - 8}{2} = \frac{z - 5}{3}\] are coplanar.


If y – 2x – k = 0 touches the conic 3x2 – 5y2 = 15, find the value of k. 


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×