मराठी

The points (1, 2, 3), (–2, 3, 4) and (7, 0, 1) are collinear. - Mathematics

Advertisements
Advertisements

प्रश्न

The points (1, 2, 3), (–2, 3, 4) and (7, 0, 1) are collinear.

पर्याय

  • True

  • False

MCQ
चूक किंवा बरोबर

उत्तर

This statement is True.

Explanation:

Let A, B, C be the points (1, 2, 3), (–2, 3, 4) and (7, 0, 1), respectively

Then, the direction ratios of each of the lines AB and BC are proportional to – 3, 1, 1.

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 11: Three Dimensional Geometry - Solved Examples [पृष्ठ २३४]

APPEARS IN

एनसीईआरटी एक्झांप्लर Mathematics [English] Class 12
पाठ 11 Three Dimensional Geometry
Solved Examples | Q 23 | पृष्ठ २३४

संबंधित प्रश्‍न

Find the equations of the planes parallel to the plane x-2y + 2z-4 = 0, which are at a unit distance from the point (1,2, 3).


Find the equation of the plane passing through the following points.

 (2, 1, 0), (3, −2, −2) and (3, 1, 7)


Find the equation of the plane passing through the following points.

 (−5, 0, −6), (−3, 10, −9) and (−2, 6, −6)


Find the equation of the plane passing through the following point

 (1, 1, 1), (1, −1, 2) and (−2, −2, 2)


Find the equation of the plane passing through the following points. 

(2, 3, 4), (−3, 5, 1) and (4, −1, 2) 

 


Find the equation of the plane passing through the following point

(0, −1, 0), (3, 3, 0) and (1, 1, 1)

 

 


Show that the four points (0, −1, −1), (4, 5, 1), (3, 9, 4) and (−4, 4, 4) are coplanar and find the equation of the common plane.


Show that the following points are coplanar.
 (0, −1, 0), (2, 1, −1), (1, 1, 1) and (3, 3, 0) 


Show that the following points are coplanar. 

 (0, 4, 3), (−1, −5, −3), (−2, −2, 1) and (1, 1, −1)

 

Find the coordinates of the point where the line through (3, -4 , -5 ) and B (2, -3 , 1) crosses the plane passing through three points L(2,2,1), M(3,0,1) and N(4, -1,0 ) . Also, find the ratio in which diveides the line segment AB.


Find the vector equations of the following planes in scalar product form  \[\left( \vec{r} \cdot \vec{n} = d \right):\] \[\vec{r} = \left( 1 + s - t \right) \hat{t}  + \left( 2 - s \right) \hat{j}  + \left( 3 - 2s + 2t \right) \hat{k} \]

 

Find the vector equations of the following planes in scalar product form  \[\left( \vec{r} \cdot \vec{n} = d \right):\]\[\vec{r} = \hat{i} - \hat{j} + \lambda\left( \hat{i}  + \hat{j}  + \hat{k}  \right) + \mu\left( 4 \hat{i}  - 2 \hat{j}  + 3 \hat{k} \right)\]

 


Find the Cartesian forms of the equations of the following planes. \[\vec{r} = \left( \hat{i}  - \hat{j}  \right) + s\left( - \hat{i}  + \hat{j}  + 2 \hat{k} \right) + t\left( \hat{i} + 2 \hat{j} + \hat{k}  \right)\]


Find the vector equation of the following planes in non-parametric form. \[\vec{r} = \left( 2 \hat{i}  + 2 \hat{j}  - \hat{k}  \right) + \lambda\left( \hat{i}  + 2 \hat{j}  + 3 \hat{k}  \right) + \mu\left( 5 \hat{i}  - 2 \hat{j} + 7 \hat{k}  \right)\]

 


Find the equation of the plane which is parallel to 2x − 3y + z = 0 and which passes through (1, −1, 2).


Find the equation of the plane through (3, 4, −1) which is parallel to the plane \[\vec{r} \cdot \left( 2 \hat{i} - 3 \hat{j}  + 5 \hat{k} \right) + 2 = 0 .\]

 

Find the equation of the plane passing through the line of intersection of the planes 2x − 7y + 4z − 3 = 0, 3x − 5y + 4z + 11 = 0 and the point (−2, 1, 3).


Find the equation of the plane passing through the points (3, 4, 1) and (0, 1, 0) and parallel to the line 

\[\frac{x + 3}{2} = \frac{y - 3}{7} = \frac{z - 2}{5} .\]
  

Show that the lines \[\vec{r} = \left( 2 \hat{j}  - 3 \hat{k} \right) + \lambda\left( \hat{i}  + 2 \hat{j}  + 3 \hat{k} \right) \text{ and } \vec{r} = \left( 2 \hat{i}  + 6 \hat{j} + 3 \hat{k} \right) + \mu\left( 2 \hat{i}  + 3 \hat{j} + 4 \hat{k}  \right)\]  are coplanar. Also, find the equation of the plane containing them.

 
 

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

 

Show that the lines  \[\frac{x + 3}{- 3} = \frac{y - 1}{1} = \frac{z - 5}{5}\] and  \[\frac{x + 1}{- 1} = \frac{y - 2}{2} = \frac{z - 5}{5}\]  are coplanar. Hence, find the equation of the plane containing these lines.

 

Find the values of  \[\lambda\] for which the lines

\[\frac{x - 1}{1} = \frac{y - 2}{2} = \frac{z + 3}{\lambda^2}\]and  \[\frac{x - 3}{1} = \frac{y - 2}{\lambda^2} = \frac{z - 1}{2}\]  are coplanar . 

If the straight lines  \[\frac{x - 1}{2} = \frac{y + 1}{k} = \frac{z}{2}\] and \[\frac{x + 1}{2} = \frac{y + 1}{2} = \frac{z}{k}\] are coplanar, find the equations of the planes containing them.

 


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×