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Karnataka Board PUCPUC Science 2nd PUC Class 12

PUC Science 2nd PUC Class 12 - Karnataka Board PUC Question Bank Solutions for Mathematics

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Mathematics
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Find the equation of the plane passing through (abc) and parallel to the plane  \[\vec{r} \cdot \left( \hat{i}  + \hat{j}  + \hat{k}  \right) = 2 .\]

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

Find the equation of the plane passing through the point (−1, 3, 2) and perpendicular to each of the planes x + 2y + 3z = 5 and 3x + 3y + z = 0.

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

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Find the vector equation of the plane through the points (2, 1, −1) and (−1, 3, 4) and perpendicular to the plane x − 2y + 4z = 10 

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

Find the equation of the plane passing through (abc) and parallel to the plane \[\vec{r} \cdot \left( \hat{i} + \hat{j} + \hat{k}  \right) = 2 .\]

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

Find the vector equation of the line through the origin which is perpendicular to the plane  \[\vec{r} \cdot \left( \hat{i} + 2 \hat{j}  + 3 \hat{k}  \right) = 3 .\]

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

Find the equation of the plane through (2, 3, −4) and (1, −1, 3) and parallel to x-axis.

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

Find the equation of a plane passing through the points (0, 0, 0) and (3, −1, 2) and parallel to the line \[\frac{x - 4}{1} = \frac{y + 3}{- 4} = \frac{z + 1}{7} .\]

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

Find the vector equation of the line passing through the point (1, −1, 2) and perpendicular to the plane 2x − y + 3z − 5 = 0.

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

Find the equation of the plane through the points (2, 2, −1) and (3, 4, 2) and parallel to the line whose direction ratios are 7, 0, 6.

 
[11] Three - Dimensional Geometry
Chapter: [11] Three - Dimensional Geometry
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Find the equation of the plane passing through the intersection of the planes x − 2y + z = 1 and 2x + y + z = 8 and parallel to the line with direction ratios proportional to 1, 2, 1. Also, find the perpendicular distance of (1, 1, 1) from this plane

[11] Three - Dimensional Geometry
Chapter: [11] Three - Dimensional Geometry
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Find the vector equation of the line passing through (1, 2, 3) and perpendicular to the plane \[\vec{r} \cdot \left( \hat{i}  + 2 \hat{j}  - 5 \hat{k}  \right) + 9 = 0 .\]

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

Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the   yz - plane .

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

Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the  zx - plane .

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

Find the coordinates of the point where the line through (3, −4, −5) and (2, −3, 1) crosses the plane 2x + y + z = 7.

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

If the lines  \[\frac{x - 1}{- 3} = \frac{y - 2}{- 2k} = \frac{z - 3}{2} \text{ and }\frac{x - 1}{k} = \frac{y - 2}{1} = \frac{z - 3}{5}\] are perpendicular, find the value of and, hence, find the equation of the plane containing these lines.

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

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

 

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

Find the image of the point (0, 0, 0) in the plane 3x + 4y − 6z + 1 = 0.

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

Find the reflection of the point (1, 2, −1) in the plane 3x − 5y + 4z = 5.

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

Find the coordinates of the foot of the perpendicular drawn from the point (5, 4, 2) to the line \[\frac{x + 1}{2} = \frac{y - 3}{3} = \frac{z - 1}{- 1} .\]

 Hence, or otherwise, deduce the length of the perpendicular.

 
 
[11] Three - Dimensional Geometry
Chapter: [11] Three - Dimensional Geometry
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Find the image of the point with position vector \[3 \hat{i} + \hat{j}  + 2 \hat{k} \]  in the plane  \[\vec{r} \cdot \left( 2 \hat{i} - \hat{j}  + \hat{k}  \right) = 4 .\]  Also, find the position vectors of the foot of the perpendicular and the equation of the perpendicular line through \[3 \hat{i}  + \hat{j}  + 2 \hat{k} .\]

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