#### Chapters

Chapter 2: Inverse Trigonometric Functions

Chapter 3: Matrices

Chapter 4: Determinants

Chapter 5: Continuity and Differentiability

Chapter 6: Application of Derivatives

Chapter 7: Integrals

Chapter 8: Application of Integrals

Chapter 9: Differential Equations

Chapter 10: Vector Algebra

Chapter 11: Three Dimensional Geometry

Chapter 12: Linear Programming

Chapter 13: Probability

#### NCERT Mathematics Class 12

## Chapter 11: Three Dimensional Geometry

#### Chapter 11: Three Dimensional Geometry solutions [Page 467]

Find the direction cosines of a line which makes equal angles with the coordinate axes.

#### Chapter 11: Three Dimensional Geometry solutions [Pages 467 - 478]

If a line makes angles 90°, 135°, 45° with *x*, *y* and *z*-axes respectively, find its direction cosines.

If a line has the direction ratios −18, 12, −4, then what are its direction cosines?

Find the equation of the line which passes through the point (1, 2, 3) and is parallel to the vector.`3hati+2hatj-2hatk`

Show that the points (2, 3, 4), (−1, −2, 1), (5, 8, 7) are collinear.

Find the equation of the line in vector and in Cartesian form that passes through the point with position vector `2hati -hatj+4hatk` and is in the direction `hati + 2hatj - hatk`.

Find the Direction Cosines of the Sides of the Triangle Whose Vertices Are (3, 5, -4), (-1, 1, 2) and (-5, -5, -2)

Find the Cartesian equation of the line which passes through the point (−2, 4, −5) and parallel to the line given by `(x+3)/3 = (y-4)/5 = ("z"+8)/6`

The Cartesian equation of a line is `(x-5)/3 = (y+4)/7 = ("z"-6)/2` Write its vector form.

The given line passes through the point (5, −4, 6). The position vector of this point is `veca = 5hati - 4hatj + 6hatk`

Also, the direction ratios of the given line are 3, 7, and 2.

This means that the line is in the direction of vector, `vecb =3hati +7hatj + 2hatk`

It is known that the line through position vector `veca` and in the direction of the vector `vecb`is given by the equation, `vecr = veca+lambdavecb, lambda in R`

This is the required equation of the given line in vector form.

Find the vector and the Cartesian equations of the lines that pass through the origin and (5, −2, 3).

Find the vector and the Cartesian equations of the line that passes through the points (3, −2, −5), (3, −2, 6).

#### Chapter 11: Three Dimensional Geometry solutions [Pages 477 - 478]

Show that the three lines with direction cosines `12/13,(-3)/13,(-4)/13; 4/13,12/13,3/13;3/13,(-4)/13,12/13 ` are mutually perpendicular.

Show that the line through the points (1, −1, 2) (3, 4, −2) is perpendicular to the line through the points (0, 3, 2) and (3, 5, 6).

Show that the line through the points (4, 7, 8) (2, 3, 4) is parallel to the line through the points (−1, −2, 1), (1, 2, 5).

Find the angle between the following pairs of lines:

`vecr = 2hati - 5hatj + hatk + lambda(3hati - 2hatj + 6hatk) and vecr = 7hati - 6hatk + mu(hati + 2hatj + 2hatk)`

Find the angle between the following pairs of lines:

`vecr = 3hati + hatj - 2hatk + lambda(hati - hatj - 2hatk) and vecr = 2hati - hatj -56hatk + mu(3hati - 5hatj - 4hatk)`

Find the angle between the following pairs of lines:

`(x-2)/2 = (y-1)/5 = (z+3)/(-3)` and `(x+2)/(-1) = (y-4)/8 = (z -5)/4`

Find the angle between the following pairs of lines:

`x/y = y/2 = z/1` and `(x-5)/4 = (y-2)/1 = (z - 3)/8`

Find the values of *p* so the line `(1-x)/3 = (7y-14)/2p = (z-3)/2` and `(7-7x)/(3p) = (y -5)/1 = (6-z)/5` are at right angles.

Show that the lines `(x-5)/7 = (y + 2)/(-5) = z/1` and `x/1 = y/2 = z/3` are perpendicular to each other.

Find the shortest distance between the lines

`vecr = (hati+2hatj+hatk) + lambda(hati-hatj+hatk)` and `vecr = 2hati - hatj - hatk + mu(2hati + hatj + 2hatk)`

Find the shortest distance between the lines `(x+1)/7 = `(y+1)/(-6) = (z+1)/1` and (x-3)/1 = (y-5)/(-2) = (z-7)/1`

Find the shortest distance between the lines whose vector equations are `vecr = (hati + 2hatj + 3hatk) + lambda(hati - 3hatj + 2hatk)` and `vecr = 4hati + 5hatj + 6hatk + mu(2hati + 3hatj + hatk)`

Find the shortest distance between the lines whose vector equations are

`vecr = (1-t)hati + (t - 2)hatj + (3 -2t)hatk` and `vecr = (s+1)hati + (2s + 1)hatk`

#### Chapter 11: Three Dimensional Geometry solutions [Pages 493 - 494]

In following cases, determine the direction cosines of the normal to the plane and the distance from the origin.

z = 2

In following cases, determine the direction cosines of the normal to the plane and the distance from the origin.

x + y + z = 1

In following cases, determine the direction cosines of the normal to the plane and the distance from the origin.

2x + 3y – z = 5

5*y* + 8 = 0

Find the vector equation of a plane which is at a distance of 7 units from the origin and normal to the vector.`3hati + 5hatj - 6hatk`

In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.

x + y + z = 1

Find the Cartesian equation of the following planes:

`vecr.(hati + hatj-hatk) = 2`

Find the Cartesian equation of the following planes:

`vecr.(2hati + 3hatj-4hatk) = 1`

Find the Cartesian equation of the following planes:

`vecr.[(s-2t)hati + (3 - t)hatj + (2s + t)hatk] = 15`

In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.

2x + 3y + 4z – 12 = 0

In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.

3y + 4z – 6 = 0

In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.

5y + 8 = 0

Find the vector and Cartesian equation of the planes that passes through the point (1, 0, −2) and the normal to the plane is `hati + hatj - hatk`

Find the vector and Cartesian equation of the planes that passes through the point (1, 4, 6) and the normal vector to the plane is `hati -2hatj + hatk`

Find the equations of the planes that passes through three points.

(1, 1, −1), (6, 4, −5), (−4, −2, 3)

Find the equations of the planes that passes through three points.

(1, 1, 0), (1, 2, 1), (−2, 2, −1)

Find the intercepts cut off by the plane 2x + y – z = 5.

Find the equation of the plane with intercept 3 on the *y*-axis and parallel to ZOX plane.

Find the equation of the plane through the intersection of the planes 3x – y + 2z – 4 = 0 and x + y + z – 2 = 0 and the point (2, 2, 1).

Find the vector equation of the plane passing through the intersection of the planes `vecr.(2hati + 2hatj - 3hatk) = 7, vecr.(2hati + 5hatj + 3hatk) = 9` and through the point (2, 1, 3)

Find the equation of the plane through the line of intersection of the planes x + y + z = 1 and 2x + 3y + 4z = 5 which is perpendicular to the plane x − y + z = 0. Also find the distance of the plane, obtained above, from the origin.

Find the angle between the planes whose vector equations are `vecr.(2hati + 2hatj - 3hatk) = 5 and hatr.(3hati - 3hatj + 5hatk) = 3`

In the following cases, determine whether the given planes are parallel or perpendicular, and in case they are neither, find the angles between them.

7x + 5y + 6z + 30 = 0 and 3x – y – 10z + 4 = 0

In the following cases, determine whether the given planes are parallel or perpendicular, and in case they are neither, find the angles between them.

2x + y + 3z – 2 = 0 and x – 2y + 5 = 0

In the following cases, determine whether the given planes are parallel or perpendicular, and in case they are neither, find the angles between them.

2x – 2y + 4z + 5 = 0 and 3x – 3y + 6z – 1 = 0

2x – y + 3z – 1 = 0 and 2x – y + 3z + 3 = 0

4x + 8y + z – 8 = 0 and y + z – 4 = 0

In the given cases, find the distance of each of the given points from the corresponding given plane.

Point Plane

(0, 0, 0) 3x – 4y + 12 z = 3

In the given cases, find the distance of each of the given points from the corresponding given plane

Point Plane

(3, – 2, 1) 2x – y + 2z + 3 = 0

In the given cases, find the distance of each of the given points from the corresponding given plane.

Point Plane

(2, 3, – 5) x + 2y – 2z = 9

In the given cases, find the distance of each of the given points from the corresponding given plane.

Point Plane

(– 6, 0, 0) 2x – 3y + 6z – 2 = 0

#### Chapter 11: Three Dimensional Geometry solutions [Pages 497 - 499]

Show that the line joining the origin to the point (2, 1, 1) is perpendicular to the line determined by the points (3, 5, – 1), (4, 3, – 1).

If *l*_{1}, *m*_{1}, *n*_{1} and *l*_{2}, *m*_{2}, *n*_{2} are the direction cosines of two mutually perpendicular lines, show that the direction cosines of the line perpendicular to both of these are *m*_{1}*n*_{2} − *m*_{2}*n*_{1}, *n*_{1}*l*_{2} − *n*_{2}*l*_{1}, *l*_{1}*m*_{2} − *l*_{2}*m*_{1}.

Find the angle between the lines whose direction ratios are *a*, *b*, *c *and *b* − *c*, *c* − *a*, *a* − *b*.

Find the equation of a line parallel to *x*-axis and passing through the origin.

If the coordinates of the points A, B, C, D be (1, 2, 3), (4, 5, 7), (−4, 3, −6) and (2, 9, 2) respectively, then find the angle between the lines AB and CD.

If the lines `(x-1)/(-3) = (y -2)/(2k) = (z-3)/2 and (x-1)/(3k) = (y-1)/1 = (z -6)/(-5)` are perpendicular, find the value of* k*.

Find the vector equation of the plane passing through (1, 2, 3) and perpendicular to the plane `vecr.(hati + 2hatj -5hatk) + 9 = 0`

Find the equation of the plane passing through (*a*, *b*, *c*) and parallel to the plane `vecr.(hati + hatj + hatk) = 2`

Find the shortest distance between lines `vecr = 6hati + 2hatj + 2hatk + lambda(hati - 2hatj + 2hatk)` and `vecr =-4hati - hatk + mu(3hati - 2hatj - 2hatk)`

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

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

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

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

If the points (1, 1, *p*) and (−3, 0, 1) be equidistant from the plane `vecr.(3hati + 4hatj - 12hatk)+ 13 = 0`, then find the value of *p*.

Find the equation of the plane passing through the line of intersection of the planes `vecr.(hati + hatj + hatk) = 1` and `vecr.(2hati + 3hatj -hatk) + 4 = 0` and parallel to *x*-axis.

If O be the origin and the coordinates of P be (1, 2, −3), then find the equation of the plane passing through P and perpendicular to OP.

Find the equation of the plane which contains the line of intersection of the planes `vecrr.(hati + 2hatj + 3hatk) - 4 = 0, vecr.(2hati + htj - hatk) + 5 = 0`, and which is perpendicular to the plane `vecr.(5hati + 3hatj - 6hatk) + 8 = 0`.

Find the distance of the point (−1, −5, −10) from the point of intersection of the line `vecr = 2hati -hatj + 2hatk + lambda(3hati + 4hatj + 2hatk)` and the plane `vecr.(hati -hatj + hatk) = 5`.

Find the vector equation of the line passing through (1, 2, 3) and parallel to the planes `vecr = (hati - hatj + 2hatk) = 5`and `vecr.(3hati + hatj + hatk) = 6`

Find the vector equation of the line passing through the point (1, 2, − 4) and perpendicular to the two lines: `(x -8)/3 = (y+19)/(-16) = (z - 10)/7 and (x - 15)/3 = (y - 29)/8 = (z- 5)/(-5)`

Prove that if a plane has the intercepts *a*, *b*, *c* and is at a distance of *P* units from the origin, then `1/a^2 + 1/b^2 + 1/c^2 = 1/p^2`

Distance between the two planes: 2x + 3y + 4z = 4 and 4x + 6y + 8z = 12 is

(A) 2 units

(B) 4 units

(C) 8 units

(D)`2/sqrt29 "units"`

The planes: 2*x *− *y* + 4*z* = 5 and 5*x* − 2.5*y* + 10*z* = 6 are

(A) Perpendicular

(B) Parallel

(C) intersect *y*-axis

(C) passes through `(0,0,5/4)`

## Chapter 11: Three Dimensional Geometry

#### NCERT Mathematics Class 12

#### Textbook solutions for Class 12

## NCERT solutions for Class 12 Mathematics chapter 11 - Three Dimensional Geometry

NCERT solutions for Class 12 Maths chapter 11 (Three Dimensional Geometry) include all questions with solution and detail explanation. This will clear students doubts about any question and improve application skills while preparing for board exams. The detailed, step-by-step solutions will help you understand the concepts better and clear your confusions, if any. Shaalaa.com has the CBSE Mathematics Textbook for Class 12 solutions in a manner that help students grasp basic concepts better and faster.

Further, we at Shaalaa.com are providing such solutions so that students can prepare for written exams. NCERT textbook solutions can be a core help for self-study and acts as a perfect self-help guidance for students.

Concepts covered in Class 12 Mathematics chapter 11 Three Dimensional Geometry are Three - Dimensional Geometry Examples and Solutions, Introduction of Three Dimensional Geometry, Equation of a Plane Passing Through Three Non Collinear Points, Relation Between Direction Ratio and Direction Cosines, Intercept Form of the Equation of a Plane, Coplanarity of Two Lines, Distance of a Point from a Plane, Angle Between Line and a Plane, Angle Between Two Planes, Angle Between Two Lines, Vector and Cartesian Equation of a Plane, Equation of a Plane in Normal Form, Equation of a Plane Perpendicular to a Given Vector and Passing Through a Given Point, Plane Passing Through the Intersection of Two Given Planes, Shortest Distance Between Two Lines, Equation of a Line in Space, Direction Cosines and Direction Ratios of a Line.

Using NCERT Class 12 solutions Three Dimensional Geometry exercise by students are an easy way to prepare for the exams, as they involve solutions arranged chapter-wise also page wise. The questions involved in NCERT Solutions are important questions that can be asked in the final exam. Maximum students of CBSE Class 12 prefer NCERT Textbook Solutions to score more in exam.

Get the free view of chapter 11 Three Dimensional Geometry Class 12 extra questions for Maths and can use Shaalaa.com to keep it handy for your exam preparation