NCERT Solutions | Class 12 Maths Chapter 11

NCERT Solutions | Class 12 Maths Chapter 11 | Three-dimensional Geometry 

CBSE Solutions | Maths Class 12

Check the below NCERT Solutions for Class 12 Maths Chapter 11 Three-dimensional Geometry Pdf free download. NCERT Solutions Class 12 Maths  were prepared based on the latest exam pattern. We have Provided Three-dimensional Geometry Class 12 Maths NCERT Solutions to help students understand the concept very well.

NCERT | Class 12 Maths

NCERT Solutions Class 12 Maths

Book:	National Council of Educational Research and Training (NCERT)
Board:	Central Board of Secondary Education (CBSE)
Class:	12th
Subject:	Maths
Chapter:	11
Chapters Name:	Three-dimensional Geometry
Medium:	English

Three-dimensional Geometry | Class 12 Maths | NCERT Books Solutions

You can refer to MCQ Questions for Class 12 Maths Chapter 11 Three-dimensional Geometry to revise the concepts in the syllabus effectively and improve your chances of securing high marks in your board exams.

NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Ex 11.1

Ex 11.1 Class 12 Maths Question 1.
If a line makes angles 90°, 135°, 45° with the and z axes respectively, find its direction cosines.
Solution.
Direction angles are 90°, 135°, 45°
Direction cosines are

Ex 11.1 Class 12 Maths Question 2.

Find the direction cosines of a line which makes equal angles with coordinate axes.
Solution.
Let direction angle be α each
∴ Direction cosines are cos α, cos α, cos α
But l² + m² + n² = 1
∴cos² a + cos² a + cos² a = 1

Ex 11.1 Class 12 Maths Question 3.
If a line has the direction ratios – 18,12, -4 then what are its direction cosines?
Solution.
Now given direction ratios of a line are -8,12,-4
∴ a = -18,b = 12,c = -4
Direction cosines are

Ex 11.1 Class 12 Maths Question 4.
Show that the points (2,3,4) (-1,-2,1), (5,8,7) are collinear.
Solution.
Let the points be A(2,3,4), B (-1, -2,1), C (5,8,7).
Let direction ratios of AB be

Ex 11.1 Class 12 Maths Question 5.
Find the direction cosines of the sides of the triangle whose vertices are (3,5, -4), (-1,1,2) and (-5,-5,-2).
Solution.
The vertices of triangle ABC are A (3, 5, -4), B (-1,1,2), C (-5, -5, -2)
(i) Direction ratios of AB are (-4,-4,6)
Direction cosines are

NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Ex 11.2

Ex 11.2 Class 12 Maths Question 1.
Show that the three lines with direction cosines:
\(\frac { 12 }{ 13 } ,\frac { -3 }{ 13 } ,\frac { -4 }{ 13 } ,\frac { 4 }{ 13 } ,\frac { 12 }{ 13 } ,\frac { 3 }{ 13 } ,\frac { 3 }{ 13 } ,\frac { -4 }{ 13 } ,\frac { 12 }{ 13 } \)
are mutually perpendicular.
Solution.
Let the lines be L₁,L₂ and L₃.
∴ For lines L1 and L2

Ex 11.2 Class 12 Maths Question 2.
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).
Solution.
Let A, B be the points (1, -1, 2), (3, 4, -2) respectively Direction ratios of AB are 2,5, -4
Let C, D be the points (0, 3, 2) and (3, 5, 6) respectively Direction ratios of CD are 3, 2,4 AB is Perpendicular to CD if

Ex 11.2 Class 12 Maths Question 3.
Show that the line through the points (4,7,8) (2,3,4) is parallel to the line through the points (-1,-2,1) and (1,2,5).
Solution.
Let the points be A(4,7,8), B (2,3,4), C (-1,-2,1) andD(1,2,5).
Now direction ratios of AB are

Ex 11.2 Class 12 Maths Question 4.
Find the equation of the line which passes through the point (1,2,3) and is parallel to the vector \(3\hat { i } +2\hat { j } -2\hat { k } \)
Solution.
Equation of the line passing through the point

Ex 11.2 Class 12 Maths Question 5.
Find the equation of the line in vector and in cartesian form that passes through the point with position vector \(2\hat { i } -\hat { j } +4\hat { k } \) and is in the direction \(\hat { i } +2\hat { j } -\hat { k } \).
Solution.
The vector equation of a line passing through a point with position vector \(\overrightarrow { a } \) and parallel to the

Ex 11.2 Class 12 Maths Question 6.
Find the cartesian equation of the line which passes through the point (-2,4, -5) and parallel to the line is given by \(\frac { x+3 }{ 3 } =\frac { y-4 }{ 5 } =\frac { z+8 }{ 6 } \)
Solution.
The cartesian equation of the line passing through the point (-2,4, -5) and parallel to the

Ex 11.2 Class 12 Maths Question 7.
The cartesian equation of a line is
\(\frac { x-5 }{ 3 } =\frac { y+4 }{ 7 } =\frac { z-6 }{ 2 } \)
write its vector form.
Solution.
The cartesian equation of the line is
\(\frac { x-5 }{ 3 } =\frac { y+4 }{ 7 } =\frac { z-6 }{ 2 } \)
Clearly (i) passes through the point (5, – 4, 6) and has 3,7,2 as its direction ratios.
=> Line (i) passes through the point A with

Ex 11.2 Class 12 Maths Question 8.
Find the vector and the cartesian equations of the lines that passes through the origin and (5,-2,3).
Solution.
The line passes through point
\(\therefore \overrightarrow { a } =\overrightarrow { 0 } \)
Direction ratios of the line passing through the

Ex 11.2 Class 12 Maths Question 9.
Find the vector and cartesian equations of the line that passes through the points (3, -2, -5), (3,-2,6).
Solution.
The PQ passes through the point P(3, -2, -5)

Ex 11.2 Class 12 Maths Question 10.
Find the angle between the following pair of lines
(i) \(\overrightarrow { r } =2\hat { i } -5\hat { j } +\hat { k } +\lambda (3\hat { i } +2\hat { j } +6\hat { k } )\)
\(and\quad \overrightarrow { r } =7\hat { i } -6\hat { j } +\mu (\hat { i } +2\hat { j } +2\hat { k } )\)
(ii) \(\overrightarrow { r } =3\hat { i } +\hat { j } -2\hat { k } +\lambda (\hat { i } -\hat { j } -2\hat { k } )\)
\(\overrightarrow { r } =2\hat { i } -\hat { j } -56\hat { k } +\mu (3\hat { i } -5\hat { j } -4\hat { k } )\)
Solution.
(i) Let θ be the angle between the given lines.
The given lines are parallel to the vectors

Ex 11.2 Class 12 Maths Question 11.
Find the angle between the following pair of lines
(i) \(\frac { x-2 }{ 2 } =\frac { y-1 }{ 5 } =\frac { z+3 }{ -3 } and\frac { x+2 }{ -1 } =\frac { y-4 }{ 8 } =\frac { z-5 }{ 4 } \)
(ii) \(\frac { x }{ 2 } =\frac { y }{ 2 } =\frac { z }{ 1 } and\frac { x-5 }{ 4 } =\frac { y-2 }{ 1 } =\frac { z-3 }{ 8 } \)
Solution.
Given
(i) \(\frac { x-2 }{ 2 } =\frac { y-1 }{ 5 } =\frac { z+3 }{ -3 } and\frac { x+2 }{ -1 } =\frac { y-4 }{ 8 } =\frac { z-5 }{ 4 } \)
(ii) \(\frac { x }{ 2 } =\frac { y }{ 2 } =\frac { z }{ 1 } and\frac { x-5 }{ 4 } =\frac { y-2 }{ 1 } =\frac { z-3 }{ 8 } \)

Ex 11.2 Class 12 Maths Question 12.
Find the values of p so that the lines
\(\frac { 1-x }{ 3 } =\frac { 7y-14 }{ 2p } =\frac { z-3 }{ 2 } and\frac { 7-7x }{ 3p } =\frac { y-5 }{ 1 } =\frac { 6-z }{ 5 } \)
are at right angles
Solution.
The given equation are not in the standard form
The equation of given lines is

Ex 11.2 Class 12 Maths Question 13.
Show that the lines \(\frac { x-5 }{ 7 } =\frac { y+2 }{ -5 } =\frac { z }{ 1 } and\frac { x }{ 1 } =\frac { y }{ 2 } =\frac { z }{ 3 } \) are perpendicular to each other
Solution.
Given lines
\(\frac { x-5 }{ 7 } =\frac { y+2 }{ -5 } =\frac { z }{ 1 } \) …(i)
\(\frac { x }{ 1 } =\frac { y }{ 2 } =\frac { z }{ 3 } \) …(ii)

Ex 11.2 Class 12 Maths Question 14.
Find the shortest distance between the lines
\(\overrightarrow { r } =(\hat { i } +2\hat { j } +\hat { k } )+\lambda (\hat { i } -\hat { j } +\hat { k } )\) and
\(\overrightarrow { r } =(2\hat { i } -\hat { j } -\hat { k } )+\mu (2\hat { i } +\hat { j } +2\hat { k } )\)
Solution.
The shortest distance between the lines

Ex 11.2 Class 12 Maths Question 15.
Find the shortest distance between the lines
\(\frac { x+1 }{ 7 } =\frac { y+1 }{ -6 } =\frac { z+1 }{ 1 } and\frac { x-3 }{ 1 } =\frac { y-5 }{ -2 } =\frac { z-7 }{ 1 } \)
Solution.
Shortest distance between the lines

Ex 11.2 Class 12 Maths Question 16.
Find the distance between die lines whose vector equations are:
\(\overrightarrow { r } =(\hat { i } +2\hat { j } +3\hat { k) } +\lambda (\hat { i } -3\hat { j } +2\hat { k } )\) and
\(\overrightarrow { r } =(4\hat { i } +5\hat { j } +6\hat { k) } +\mu (2\hat { i } +3\hat { j } +\hat { k } )\)
Solution.
Comparing the given equations with

Ex 11.2 Class 12 Maths Question 17.
Find the shortest distance between the lines whose vector equations are
\(\overrightarrow { r } =(1-t)\hat { i } +(t-2)\hat { j } +(3-2t)\hat { k } \) and
\(\overrightarrow { r } =(s+1)\hat { i } +(2s-1)\hat { j } -(2s+1)\hat { k } \)
Solution.
Comparing these equation with

NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Ex 11.3

Ex 11.3 Class 12 Maths Question 1.
In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.
(a) z = 2
(b) x+y+z = 1
(c) 2x + 3y – z = 5
(d) 5y+8 = 0
Solution.
(a) Direction ratios of the normal to the plane are 0,0,1
=> a = 0, b = 0, c = 1

<

Ex 11.3 Class 12 Maths Question 2.
Find the vector equation of a plane which is at a distance of 7 units from the origin and normal to the vector \(3\hat { i } +5\hat { j } -6\hat { k } \)
Solution.

Ex 11.3 Class 12 Maths Question 3.
Find the Cartesian equation of the following planes.
(a) \(\overrightarrow { r } \cdot (\hat { i } +\hat { j } -\hat { k) } =2\)
(b) \(\overrightarrow { r } \cdot (\hat { 2i } +3\hat { j } -4\hat { k) } =1\)
(c) \(\overrightarrow { r } \cdot [(s-2t)\hat { i } +(3-t)\hat { j } +(2s+t)\hat { k) } =15\)
Solution.
(a) \(\overrightarrow { r } \) is the position vector of any arbitrary point P (x, y, z) on the plane.

Ex 11.3 Class 12 Maths Question 4.
In the following cases find the coordinates of the foot of perpendicular drawn from the origin
(a) 2x + 3y + 4z – 12 = 0
(b) 3y + 4z – 6 = 0
(c) x + y + z = 1
(d) 5y + 8 = 0
Solution.
(a) Let N (x1, y1, z1) be the foot of the perpendicular from the origin to the plane 2x+3y+4z-12 = 0
∴ Direction ratios of the normal are 2, 3, 4.
Also the direction ratios of ON are (x1,y1,z1)

Ex 11.3 Class 12 Maths Question 5.
Find the vector and cartesian equation of the planes
(a) that passes through the point (1,0, -2) and the normal to the plane is \(\hat { i } +\hat { j } -\hat { k } \)
(b) that passes through the point (1,4,6) and the normal vector to the plane is \(\hat { i } -2\hat { j } +\hat { k } \)
Solution.
(a) Normal to the plane is i + j – k and passes through (1,0,-2)

Ex 11.3 Class 12 Maths Question 6.
Find the equations of the planes that passes through three points
(a) (1,1,-1) (6,4,-5), (-4, -2,3)
(b) (1,1,0), (1,2,1), (-2,2,-1)
Solution.
(a) The plane passes through the points (1,1,-1) (6,4,-5), (-4,-2,3)
Let the equation of the plane passing through(1,1,-1)be

Ex 11.3 Class 12 Maths Question 7.
Find the intercepts cut off by the plane 2x+y-z = 5.
Solution.
Equation of the plane is 2x + y- z = 5 x y z
Dividing by 5: \(\Rightarrow \frac { x }{ \frac { 5 }{ 2 } } +\frac { y }{ 5 } -\frac { z }{ -5 } =1\)
∴ The intercepts on the axes OX, OY, OZ are \(\frac { 5 }{ 2 }\), 5, -5 respectively

Ex 11.3 Class 12 Maths Question 8.
Find the equation of the plane with intercept 3 on the y- axis and parallel to ZOX plane.
Solution.
Any plane parallel to ZOX plane is y=b where b is the intercept on y-axis.
∴ b = 3.
Hence equation of the required plane is y = 3.

Ex 11.3 Class 12 Maths Question 9.
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).
Solution.
Given planes are:
3x – y + 2z – 4 = 0 and x + y + z – 2 = 0
Any plane through their intersection is
3x – y + 2z – 4 + λ(x + y + z – 2) = 0
point (2,2,1) lies on it,
∴3 x 2 – 2 + 2 x 1 – 4 +λ(2+2+1-2)=0
=>λ = \(\frac { -2 }{ 3 }\)
Now required equation is 7x – 5y + 4z – 8 = 0

Ex 11.3 Class 12 Maths Question 10.
Find the vector equation of the plane passing through the intersection of the planes \(\overrightarrow { r } \cdot \left( 2\hat { i } +2\hat { j } -3\hat { k } \right) =7,\overrightarrow { r } \cdot \left( 2\hat { i } +5\hat { j } +3\hat { k } \right) =9\) and through the point (2,1,3).
Solution.
Equation of the plane passing through the line of intersection of the planes

Ex 11.3 Class 12 Maths Question 11.
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.
Solution.
Given planes are
x + y + z – 1 = 0 …(i)
2x + 3y + 4z – 5 = 0 …(ii)
x – y + z = 0 ….(iii)

Ex 11.3 Class 12 Maths Question 12.
Find the angle between the planes whose vector equations are \(\overrightarrow { r } \cdot \left( 2\hat { i } +2\hat { j } -3\hat { k } \right) =5,\overrightarrow { r } \cdot \left( 3\hat { i } -3\hat { j } +5\hat { k } \right) =3\)
Solution.
The angle θ between the given planes is

Ex 11.3 Class 12 Maths Question 13.
In the following determine whether the given planes are parallel or perpendicular, and in case they are neither, find the angle between them.
(a) 7x + 5y + 6z + 30 = 0 and 3x – y – 10z + 4 = 0
(b) 2x + y + 3z – 2 = 0 and x – 2y + 5 = 0
(c) 2x – 2y + 4z + 5 = 0 and 3x – 3y + 6z – 1 = 0
(d) 2x – y + 3z – 1 = 0 and 2x – y + 3z + 3 = 0
(e) 4x + 8y + z – 8 = 0 and y + z – 4 = 0.
Solution.
(a) Direction ratios of the normal of the planes 7x + 5y + 6z + 30 = 0 are 7,5,6
Direction ratios of the normal of the plane 3x – y – 10z + 4 = 0 are 3,-1,-10
The plane 7x + 5y + 6z + 30 = 0 …(i)
3x – y – 10z + y = 0 …(ii)

Ex 11.3 Class 12 Maths Question 14.
In the following cases, find the distance of each of the given points from the corresponding given plane.
Point Plane
(a) (0, 0,0) 3x – 4y + 12z = 3
(b) (3,-2,1) 2x – y + 2z + 3 = 0.
(c) (2,3,-5) x + 2y – 2z = 9
(d) (-6,0,0) 2x – 3y + 6z – 2 = 0
Solution.
(a) Given plane: 3x – 4y + 12z – 3 = 0

NCERT Class 12 Maths

Class 12 Maths Chapters | Maths Class 12 Chapter 11

Chapterwise NCERT Solutions for Class 12 Maths

NCERT Solutions For Class 12 Maths Chapter 1 Relations and Functions
NCERT Solutions For Class 12 Maths Chapter 2 Inverse Trigonometric Functions
NCERT Solutions For Class 12 Maths Chapter 3 Matrix
NCERT Solutions For Class 12 Maths Chapter 4 Determinants
NCERT Solutions For Class 12 Maths Chapter 5 Continuity and Differentiability
NCERT Solutions For Class 12 Maths Chapter 6 Application of Derivatives
NCERT Solutions For Class 12 Maths Chapter 7 Integrals
NCERT Solutions For Class 12 Maths Chapter 8 Application of Integrals
NCERT Solutions For Class 12 Maths Chapter 9 Differential Equations
NCERT Solutions For Class 12 Maths Chapter 10 Vector Algebra
NCERT Solutions For Class 12 Maths Chapter 11 Three-dimensional Geometry
NCERT Solutions For Class 12 Maths Chapter 12 Linear Programming
NCERT Solutions For Class 12 Maths Chapter 13 Probability

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