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

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Mathematics
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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
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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

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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
Concept: undefined >> undefined

Write the normal form of the equation of the plane 2x − 3y + 6z + 14 = 0.

 
[11] Three - Dimensional Geometry
Chapter: [11] Three - Dimensional Geometry
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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
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Find the distance of the plane 2x − 3y + 4z − 6 = 0 from the origin.

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

Write the vector equation of the line passing through the point (1, −2, −3) and normal to the plane \[\vec{r} \cdot \left( 2 \hat{i} + \hat{j}  + 2 \hat{k}  \right) = 5 .\]

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

Find the vector equation of a plane which is at a distance of 5 units from the origin and its normal vector is \[2 \hat{i} - 3 \hat{j} + 6 \hat{k} \] .

[11] Three - Dimensional Geometry
Chapter: [11] Three - Dimensional Geometry
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The equation of the plane containing the two lines

\[\frac{x - 1}{2} = \frac{y + 1}{- 1} = \frac{z - 0}{3} \text{ and }\frac{x}{- 2} = \frac{y - 2}{- 3} = \frac{z + 1}{- 1}\]
 
 
[11] Three - Dimensional Geometry
Chapter: [11] Three - Dimensional Geometry
Concept: undefined >> undefined
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