मराठी

Commerce (English Medium) इयत्ता १२ - CBSE Question Bank Solutions for Mathematics

Advertisements
विषय
अध्याय
विषय
मुख्य विषय
अध्याय
Advertisements
Advertisements
Mathematics
< prev  1261 to 1280 of 4674  next > 

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.

[11] Three - Dimensional Geometry
Chapter: [11] Three - Dimensional Geometry
Concept: undefined >> undefined

Prove that the lines through A (0, −1, −1) and B (4, 5, 1) intersects the line through C (3, 9, 4) and D (−4, 4, 4). Also, find their point of intersection. 

[11] Three - Dimensional Geometry
Chapter: [11] Three - Dimensional Geometry
Concept: undefined >> undefined

Advertisements

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.

[11] Three - Dimensional Geometry
Chapter: [11] Three - Dimensional Geometry
Concept: undefined >> undefined

Determine whether the following pair of lines intersect or not: 

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

[11] Three - Dimensional Geometry
Chapter: [11] Three - Dimensional Geometry
Concept: undefined >> undefined

Determine whether the following pair of lines intersect or not: 

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

[11] Three - Dimensional Geometry
Chapter: [11] Three - Dimensional Geometry
Concept: undefined >> undefined

Determine whether the following pair of lines intersect or not: 

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

[11] Three - Dimensional Geometry
Chapter: [11] Three - Dimensional Geometry
Concept: undefined >> undefined

Determine whether the following pair of lines intersect or not:  

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

[11] Three - Dimensional Geometry
Chapter: [11] Three - Dimensional Geometry
Concept: undefined >> undefined

Show that the lines \[\vec{r} = 3 \hat{i} + 2 \hat{j} - 4 \hat{k} + \lambda\left( \hat{i} + 2 \hat{j} + 2 \hat{k} \right) \text{ and } \vec{r} = 5 \hat{i} - 2 \hat{j}  + \mu\left( 3 \hat{i} + 2 \hat{j} + 6 \hat{k} \right)\] are intersecting. Hence, find their point of intersection.

[11] Three - Dimensional Geometry
Chapter: [11] Three - Dimensional Geometry
Concept: undefined >> undefined

Find the perpendicular distance of the point (3, −1, 11) from the line \[\frac{x}{2} = \frac{y - 2}{- 3} = \frac{z - 3}{4} .\]

[11] Three - Dimensional Geometry
Chapter: [11] Three - Dimensional Geometry
Concept: undefined >> undefined

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.

[11] Three - Dimensional Geometry
Chapter: [11] Three - Dimensional Geometry
Concept: undefined >> undefined

Find the foot of the perpendicular drawn from the point A (1, 0, 3) to the joint of the points B (4, 7, 1) and C (3, 5, 3). 

[11] Three - Dimensional Geometry
Chapter: [11] Three - Dimensional Geometry
Concept: undefined >> undefined

A (1, 0, 4), B (0, −11, 3), C (2, −3, 1) are three points and D is the foot of perpendicular from A on BC. Find the coordinates of D

[11] Three - Dimensional Geometry
Chapter: [11] Three - Dimensional Geometry
Concept: undefined >> undefined

Find the foot of perpendicular from the point (2, 3, 4) to the line \[\frac{4 - x}{2} = \frac{y}{6} = \frac{1 - z}{3} .\] Also, find the perpendicular distance from the given point to the line.

[11] Three - Dimensional Geometry
Chapter: [11] Three - Dimensional Geometry
Concept: undefined >> undefined

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

[11] Three - Dimensional Geometry
Chapter: [11] Three - Dimensional Geometry
Concept: undefined >> undefined

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

[11] Three - Dimensional Geometry
Chapter: [11] Three - Dimensional Geometry
Concept: undefined >> undefined

Find the foot of the perpendicular drawn from the point  \[\hat{i} + 6 \hat{j} + 3 \hat{k} \]  to the line  \[\overrightarrow{r} = \hat{j} + 2 \hat{k} + \lambda\left( \hat{i} + 2 \hat{j} + 3 \hat{k}  \right) .\]  Also, find the length of the perpendicular

[11] Three - Dimensional Geometry
Chapter: [11] Three - Dimensional Geometry
Concept: undefined >> undefined

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.

[11] Three - Dimensional Geometry
Chapter: [11] Three - Dimensional Geometry
Concept: undefined >> undefined

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

[11] Three - Dimensional Geometry
Chapter: [11] Three - Dimensional Geometry
Concept: undefined >> undefined

Find the foot of the perpendicular from (1, 2, −3) to the line \[\frac{x + 1}{2} = \frac{y - 3}{- 2} = \frac{z}{- 1} .\]

[11] Three - Dimensional Geometry
Chapter: [11] Three - Dimensional Geometry
Concept: undefined >> undefined

Find the equation of line passing through the points A (0, 6, −9) and B (−3, −6, 3). If D is the foot of perpendicular drawn from a point C (7, 4, −1) on the line AB, then find the coordinates of the point D and the equation of line CD

[11] Three - Dimensional Geometry
Chapter: [11] Three - Dimensional Geometry
Concept: undefined >> undefined
< prev  1261 to 1280 of 4674  next > 
Advertisements
Advertisements
CBSE Commerce (English Medium) इयत्ता १२ Question Bank Solutions
Question Bank Solutions for CBSE Commerce (English Medium) इयत्ता १२ Accountancy
Question Bank Solutions for CBSE Commerce (English Medium) इयत्ता १२ Business Studies
Question Bank Solutions for CBSE Commerce (English Medium) इयत्ता १२ Computer Science (Python)
Question Bank Solutions for CBSE Commerce (English Medium) इयत्ता १२ Economics
Question Bank Solutions for CBSE Commerce (English Medium) इयत्ता १२ English Core
Question Bank Solutions for CBSE Commerce (English Medium) इयत्ता १२ English Elective - NCERT
Question Bank Solutions for CBSE Commerce (English Medium) इयत्ता १२ Entrepreneurship
Question Bank Solutions for CBSE Commerce (English Medium) इयत्ता १२ Geography
Question Bank Solutions for CBSE Commerce (English Medium) इयत्ता १२ Hindi (Core)
Question Bank Solutions for CBSE Commerce (English Medium) इयत्ता १२ Hindi (Elective)
Question Bank Solutions for CBSE Commerce (English Medium) इयत्ता १२ History
Question Bank Solutions for CBSE Commerce (English Medium) इयत्ता १२ Informatics Practices
Question Bank Solutions for CBSE Commerce (English Medium) इयत्ता १२ Mathematics
Question Bank Solutions for CBSE Commerce (English Medium) इयत्ता १२ Physical Education
Question Bank Solutions for CBSE Commerce (English Medium) इयत्ता १२ Political Science
Question Bank Solutions for CBSE Commerce (English Medium) इयत्ता १२ Psychology
Question Bank Solutions for CBSE Commerce (English Medium) इयत्ता १२ Sanskrit (Core)
Question Bank Solutions for CBSE Commerce (English Medium) इयत्ता १२ Sanskrit (Elective)
Question Bank Solutions for CBSE Commerce (English Medium) इयत्ता १२ Sociology
Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×