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Arts (English Medium) कक्षा १२ - CBSE Question Bank Solutions for Mathematics

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Mathematics
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Find the equation of the plane passing through the point (2, 3, 1), given that the direction ratios of the normal to the plane are proportional to 5, 3, 2.

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

If the axes are rectangular and P is the point (2, 3, −1), find the equation of the plane through P at right angles to OP.

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

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Find the intercepts made on the coordinate axes by the plane 2x + y − 2z = 3 and also find the direction cosines of the normal to the plane.

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

Reduce the equation 2x − 3y − 6z = 14 to the normal form and, hence, find the length of the perpendicular from the origin to the plane. Also, find the direction cosines of the normal to the plane. 

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

Reduce the equation \[\vec{r} \cdot \left( \hat{i}  - 2 \hat{j}  + 2 \hat{k}  \right) + 6 = 0\] to normal form and, hence, find the length of the perpendicular from the origin to the plane.

 

[11] Three - Dimensional Geometry
Chapter: [11] Three - Dimensional Geometry
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Write the normal form of the equation of the plane 2x − 3y + 6z + 14 = 0.

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

The direction ratios of the perpendicular from the origin to a plane are 12, −3, 4 and the length of the perpendicular is 5. Find the equation of the plane. 

[11] Three - Dimensional Geometry
Chapter: [11] Three - Dimensional Geometry
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Find a unit normal vector to the plane x + 2y + 3z − 6 = 0.

 
[11] Three - Dimensional Geometry
Chapter: [11] Three - Dimensional Geometry
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Find the equation of a plane which is at a distance of \[3\sqrt{3}\]  units from the origin and the normal to which is equally inclined to the coordinate axes.

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

Find the vector equation of the plane which is at a distance of \[\frac{6}{\sqrt{29}}\] from the origin and its normal vector from the origin is  \[2 \hat{i} - 3 \hat{j} + 4 \hat{k} .\] Also, find its Cartesian form. 

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

Find the distance of the plane 2x − 3y + 4z − 6 = 0 from the origin.

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

Find the equation of the plane which contains the line of intersection of the planes \[x + 2y + 3z - 4 = 0 \text { and } 2x + y - z + 5 = 0\] and whose x-intercept is twice its z-intercept.

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

Prove that the line of section of the planes 5x + 2y − 4z + 2 = 0 and 2x + 8y + 2z − 1 = 0 is parallel to the plane 4x − 2y − 5z − 2 = 0.

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

Find the value of λ such that the line \[\frac{x - 2}{6} = \frac{y - 1}{\lambda} = \frac{z + 5}{- 4}\]  is perpendicular to the plane 3x − y − 2z = 7.

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

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

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

Find the vector equation of the plane passing through the points (3, 4, 2) and (7, 0, 6) and perpendicular to the plane 2x − 5y − 15 = 0. Also, show that the plane thus obtained contains the line \[\vec{r} = \hat{i} + 3 \hat{j}  - 2 \hat{k}  + \lambda\left( \hat{i}  - \hat{j}  + \hat{k}  \right) .\]

 
[11] Three - Dimensional Geometry
Chapter: [11] Three - Dimensional Geometry
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Two schools P and Q want to award their selected students on the values of tolerance, kindness and leadership. School P wants to award Rs x each, Rs y each and Rs z each for the three respective values to 3, 2 and 1 students, respectively, with a total award money of Rs 2,200. School Q wants to spend Rs 3,100 to award 4, 1 and 3 students on the respective values (by giving the same award money to the three values as school P). If the total amount of award for one prize on each value is Rs 1,200, using matrices, find the award money for each value.

[3] Matrices
Chapter: [3] Matrices
Concept: undefined >> undefined

Write the plane  \[\vec{r} \cdot \left( 2 \hat{i}  + 3 \hat{j}  - 6 \hat{k}  \right) = 14\]  in normal form.

 
 
[11] Three - Dimensional Geometry
Chapter: [11] Three - Dimensional Geometry
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Write a vector normal to the plane  \[\vec{r} = l \vec{b} + m \vec{c} .\]

 
[11] Three - Dimensional Geometry
Chapter: [11] Three - Dimensional Geometry
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Write the value of k for which the line \[\frac{x - 1}{2} = \frac{y - 1}{3} = \frac{z - 1}{k}\]  is perpendicular to the normal to the plane  \[\vec{r} \cdot \left( 2 \hat{i}  + 3 \hat{j}  + 4 \hat{k}  \right) = 4 .\]

[11] Three - Dimensional Geometry
Chapter: [11] Three - Dimensional Geometry
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