NCERT Solutions for Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry

Edited By Ramraj Saini | Updated on Sep 25, 2023 07:33 PM IST

Introduction to Three Dimensional Geometry Class 11 Questions And Answers

NCERT solutions for class 11 maths chapter 12 Introduction to Three Dimensional Geometry discussed here. In our previous classes, you have studied the basic concepts of geometry in two-dimensional space. In this article, you will get introduction to three dimensional geometry class 11 NCERT solutions. These NCERT solutions are prepared by subject matter expert at Careers360 keeping in my the latest CBSE syllabus 2023.

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Introduction to Three Dimensional Geometry Class 11 Questions And Answers
Introduction to Three Dimensional Geometry Class 11 Questions And Answers PDF Free Download
Introduction to Three Dimensional Geometry Class 11 Solutions - Important Formulae
Introduction to Three Dimensional Geometry Class 11 NCERT Solutions (Intext Questions and Exercise)
NCERT Solutions for Class 11 Mathematics
Key Features Of Introduction to Three Dimensional Geometry Class 11 Solutions
NCERT Solutions for Class 11 - Subject wise
NCERT Books and NCERT Syllabus

NCERT Solutions for Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry

In introduction to three Dimensional geometry class 11 chapter, you will learn what is 3 D coordinate geometry, how a coordinate space can be determined, about coordinates of a point in space, the distance between two points and section formula. There are 14 questions in 3 exercises in this chapter. First, try to solve solve the problem on your own. If you are not able to do, you can take the help of NCERT solutions for class 11 maths chapter 12 which is prepared in a detailed manner. Students can find NCERT solutions for class 11 here.

Introduction to Three Dimensional Geometry Class 11 Questions And Answers PDF Free Download

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Introduction to Three Dimensional Geometry Class 11 Solutions - Important Formulae

Coordinate Axes in Three Dimensions:

In three dimensions, a rectangular cartesian coordinate system has three mutually perpendicular axes: X, Y, and Z.

Coordinate Planes:

The pair of axes determines three coordinate planes: XY, YZ, and ZX planes. These planes divide space into eight regions called octants.

Coordinates of a Point in Space:

The coordinates of a point P in three-dimensional space are represented as a triplet (x, y, z), which corresponds to the perpendicular distances from P on three mutually perpendicular coordinate planes: YZ-plane, ZX-plane, and XY-plane.

Coordinates of points on coordinate axes are:

X-axis: (x, 0, 0)
Y-axis: (0, y, 0)
Z-axis: (0, 0, z)

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Distance Formula in 3D Space:

The distance between two points A(x₁, y₁, z₁) and B(x₂, y₂, z₂) is given by:

AB = √((x₂-x₁)² + (y₂-y₁)² + (z₂-z₁)²)

Distance from Origin in 3D Space:

The distance of a point A(x, y, z) from the origin O(0, 0, 0) is given by:

OA = √(x² + y² + z²)

Section Formula for Internal and External Division in 3D Space:

The coordinates of a point R that divides the line segment joining two points P(x₁, y₁, z₁) and Q(x₂, y₂, z₂) internally or externally in the ratio m : n are given by:

Internal Division: ((mx₂ + nx₁)/(m + n), (my₂ + ny₁)/(m + n), (mz₂ + nz₁)/(m + n))
External Division: ((mx₂ - nx₁)/(m - n), (my₂ - ny₁)/(m - n), (mz₂ - nz₁)/(m - n))

Midpoint Formula in 3D Space:

The midpoint of the line segment joining two points (x₁, y₁, z₁) and (x₂, y₂, z₂) is given by:

((x₁ + x₂)/2, (y₁ + y₂)/2, (z₁ + z₂)/2)

Coordinates of Centroid of a Triangle in 3D Space:

For a triangle with vertices (x₁, y₁, z₁), (x₂, y₂, z₂), and (x₃, y₃, z₃), the coordinates of the centroid are given by:

((x₁ + x₂ + x₃)/3, (y₁ + y₂ + y₃)/3, (z₁ + z₂ + z₃)/3)

Free download NCERT Solutions for Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry for CBSE Exam.

Introduction to Three Dimensional Geometry Class 11 NCERT Solutions (Intext Questions and Exercise)

Class 11 maths chapter 12 question answer - Exercise: 12.1

Question:1 A point is on the x-axis. What are its y-coordinate and z-coordinates?

Answer:

Any point on x-axis have zero y coordinate and zero z coordinate.

Question:2 A point is in the XZ-plane. What can you say about its y-coordinate?

Answer:

When a point is in XZ plane, the y coordinate of this point will always be zero.

Question:3 Name the octants in which the following points lie:
(1, 2, 3), (4, –2, 3), (4, –2, –5), (4, 2, –5), (– 4, 2, –5), (– 4, 2, 5), (–3, –1, 6) (– 2, – 4, –7).

Answer:

The x-coordinate, y-coordinate, and z-coordinate of point (1, 2, 3) are all positive. Therefore, this point lies in octant I.

The x-coordinate, y-coordinate, and z-coordinate of point (4, -2, 3) are positive, negative, and positive respectively. Therefore, this point lies in octant IV.

The x-coordinate, y-coordinate, and z-coordinate of point (4, -2, -5) are positive, negative, and negative respectively. Therefore, this point lies in octant VIII.

The x-coordinate, y-coordinate, and z-coordinate of point (4, 2, -5) are positive, positive, and negative respectively. Therefore, this point lies in octant V.

The x-coordinate, y-coordinate, and z-coordinate of point (-4, 2, -5) are negative, positive, and negative respectively. Therefore, this point lies in octant VI.

The x-coordinate, y-coordinate, and z-coordinate of point (-4, 2, 5) are negative, positive, and positive respectively. Therefore, this point lies in octant II.

The x-coordinate, y-coordinate, and z-coordinate of point (-3, -1, 6) are negative, negative, and positive respectively. Therefore, this point lies in octant III.

The x-coordinate, y-coordinate, and z-coordinate of point (2, -4, -7) are positive, negative, and negative respectively. Therefore, this point lies in octant VIII.

Question:4 (i) Fill in the blanks: The x-axis and y-axis taken together determine a plane known as_______.

Answer:

The x-axis and y-axis taken together determine a plane known as XY Plane .

Question:4 (ii) Fill in the blanks: The coordinates of points in the XY-plane are of the form _______.

Answer:

The coordinates of points in the XY-plane are of the form (x, y, 0 ) .

Question:4 (iii) Fill in the blanks: Coordinate planes divide the space into ______ octants.

Answer:

Coordinate planes divide the space into Eight octants.

Class 11 maths chapter 12 question answer - Exercise: 12.2

Question:1 (i) Find the distance between the following pairs of points: (2, 3, 5) and (4, 3, 1)

Answer:

The distance between two points $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ is given by

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$

So, distance between (2, 3, 5) and (4, 3, 1) is given by

$d=\sqrt{(4-2)^2+(3-3)^2+(1-5)^2}$

$d=\sqrt{4+0+16}$

$d=\sqrt{20}$

$d=2\sqrt{5}$

Question:1 (ii) Find the distance between the following pairs of points: (–3, 7, 2) and (2, 4, –1)

Answer:

The distance between two points $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ is given by

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$

So, distance between (–3, 7, 2) and (2, 4, –1) is given by

$d=\sqrt{(2-(-3))^2+(4-7)^2+(-1-2)^2}$

$d=\sqrt{25+9+9}$

$d=\sqrt{43}$

Question:1 (iii) Find the distance between the following pairs of points:(–1, 3, – 4) and (1, –3, 4)

Answer:

The distance between two points $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ is given by

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$

So, distance between (–1, 3, – 4) and (1, –3, 4) is given by

$d=\sqrt{(1-(-1))^2+(-3-3)^2+(4-(-4))^2}$

$d=\sqrt{4+36+64}$

$d=\sqrt{104}$

$d=2\sqrt{26}$

Question:1 (iv) Find the distance between the following pairs of points: (2, –1, 3) and (–2, 1, 3).

Answer:

The distance between two points $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ is given by

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$

So, distance between (2, –1, 3) and (–2, 1, 3).) is given by

$d=\sqrt{(-2-(-2))^2+(1-(-1))^2+(3-3)^2}$

$d=\sqrt{16+4+0}$

$d=\sqrt{20}$

$d=2\sqrt{5}$

Question:2 Show that the points (–2, 3, 5), (1, 2, 3) and (7, 0, –1) are collinear.

Answer:

Given,theree points A=(–2, 3, 5), B=(1, 2, 3) and C=(7, 0, –1)

The distance between two points $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ is given by

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$

The distance AB :

$AB=\sqrt{(1-(-2))^2+(2-3)^2+(3-5)^2}$

$AB=\sqrt{(3)^2+(-1)^2+(-2)^2}$

$AB=\sqrt{9+1+4}$

$AB=\sqrt{14}$

The distance BC:

$BC=\sqrt{(7-1)^2+(0-2)^2+(-1-3)^2}$

$BC=\sqrt{36+4+16}$

$BC=\sqrt{56}$

$BC=2\sqrt{14}$

The distance CA

$CA=\sqrt{(7-(-2))^2+(0-3)^2+(-1-5)^2}$

$CA=\sqrt{81+9+36}$

$CA=\sqrt{126}$

$CA=3\sqrt{14}$

As we can see here,

$AB+BC=\sqrt{14}+2\sqrt{14}=3\sqrt{14}=AC$

Hence we can say that point A,B and C are colinear.

Question:3 (i) Verify the following: (0, 7, –10), (1, 6, – 6) and (4, 9, – 6) are the vertices of an isosceles triangle.

Answer:

Given Three points A=(0, 7, –10), B=(1, 6, – 6) and C=(4, 9, – 6)

The distance between two points $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ is given by

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$

The distance AB

$AB=\sqrt{(1-0)^2+(6-7)^2+(-6-(-10))^2}$

$AB=\sqrt{1+1+16}$

$AB=\sqrt{18}$

The distance BC

$BC=\sqrt{(4-1)^2+(9-6)^2+(-6-(-6))^2}$

$BC=\sqrt{9+9+0}$

$BC=\sqrt{18}$

The distance CA

$CA=\sqrt{(4-0)^2+(9-7)^2+(-6-(-10))^2}$

$CA=\sqrt{16+4+16}$

$CA=\sqrt{36}$

$CA=6$

As we can see $AB=BC\neq CA$

Hence we can say that ABC is an isosceles triangle.

Question:3(ii) Verify the following

(0, 7, 10), (- 1, 6, 6) and (- 4, 9, 6) are the vertices of right angled triangle.

Answer:

Given Three points A=(0, 7, 10), B=(-1, 6, 6) and C=(-4, 9, 6)

The distance between two points $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ is given by

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$

The distance AB

$AB=\sqrt{(-1-0)^2+(6-7)^2+(6-10)^2}$

$AB=\sqrt{1+1+16}$

$AB=\sqrt{18}$

The distance BC

$BC=\sqrt{(-4-(-1))^2+(9-6)^2+(6-6)^2}$

$BC=\sqrt{9+9+0}$

$BC=\sqrt{18}$

The distance CA

$CA=\sqrt{(-4-0)^2+(9-7)^2+(6-10)^2}$

$CA=\sqrt{16+4+16}$

$CA=\sqrt{36}$

$CA=6$

As we can see

$(AB)^2+(BC)^2=18+18=36=(CA)^2$

Since this follows pythagorus theorem, we can say that ABC is a right angle triangle.

Question:3(iii) Verify the following: (–1, 2, 1), (1, –2, 5), (4, –7, 8) and (2, –3, 4) are the vertices of a parallelogram.

Answer:

Given A=(–1, 2, 1), B=(1, –2, 5), C=(4, –7, 8) and D=(2, –3, 4)

Given Three points A=(0, 7, –10), B=(1, 6, – 6) and C=(4, 9, – 6)

The distance between two points $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ is given by

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$

The distance AB

$AB=\sqrt{(1-(-1))^2+(-2-2)^2+(5-1)^2}$

$AB=\sqrt{4+16+16}$

$AB=\sqrt{36}$

$AB=6$

The distance BC

$BC=\sqrt{(4-1)^2+(-7-(-2))^2+(8-5)^2}$

$BC=\sqrt{9+25+9}$

$BC=\sqrt{43}$

The distance CD

$CD=\sqrt{(2-4)^2+(-3-(-7))^2+(4-8)^2}$

$CA=\sqrt{4+16+16}$

$CA=\sqrt{36}$

$CA=6$

The distance DA

$DA=\sqrt{(-1-2)^2+(2-(-3))^2+(1-4)^2}$

$DA=\sqrt{9+25+9}$

$DA=\sqrt{43}$

Here As we can see

$AB=6=CA$ And $BC=\sqrt{43}=DA$

As the opposite sides of quadrilateral are equal, we can say that ABCD is a parallelogram.

Question:4 Find the equation of the set of points which are equidistant from the points (1, 2, 3) and (3, 2, –1).

Answer:

Given, two points A=(1, 2, 3) and B=(3, 2, –1).

Let the point P= (x,y,z) be a point which is equidistance from the points A and B.

so,

The distance PA= The distance PB

$\sqrt{(x-1)^2+(y-2)^2+(z-3)^2}=\sqrt{(x-3)^2+(y-2)^2+(z-(-1))^2}$

${(x-1)^2+(y-2)^2+(z-3)^2}={(x-3)^2+(y-2)^2+(z-(-1))^2}$

$\left [ (x-1)^2-(x-3)^2 \right ]+\left [ (y-2)^2-(y-2)^2 \right ]+\left [ (z-3)^2-(z+1)^2 \right ]=0$

Now lets apply the simplification property,

$a^2-b^2=(a+b)(a-b)$

$\left [ (2)(2x-4) \right ]+0+\left [ (-4)(2z-2) \right ]=0$

$4x-8-8z+8=0$

$4x-8z=0$

$x-2z=0$

Hence locus of the point which is equidistant from A and B is $x-2z=0$ .

Question:5 Find the equation of the set of points P, the sum of whose distances from A (4, 0, 0) and B (– 4, 0, 0) is equal to 10

Answer:

Given,

Two points A (4, 0, 0) and B (– 4, 0, 0)

let the point P(x,y,z) be a point sum of whose distance from A and B is 10.

So,

The distance PA+The distance PB=10

$\sqrt{(x-4)^2+(y-0)^2+(z-0)^2}+\sqrt{(x-(-4))^2+(y)^2+(z)^2}=10$

$\sqrt{(x-4)^2+(y)^2+(z)^2}+\sqrt{(x+4)^2+(y)^2+(z)^2}=10$

$\sqrt{(x-4)^2+(y)^2+(z)^2}=10-\sqrt{(x+4)^2+(y)^2+(z)^2}$

Squaring on both side :

${(x-4)^2+(y)^2+(z)^2}=100-20\sqrt{(x+4)^2+(y)^2+(z)^2}+{(x+4)^2+(y)^2+(z)^2}$

${(x-4)^2-(x+4)^2=100-20\sqrt{(x+4)^2+(y)^2+(z)^2}$

$-16x=100-20\sqrt{(x+4)^2+(y)^2+(z)^2}$

$20\sqrt{(x+4)^2+(y)^2+(z)^2}=100+16x$

$5\sqrt{(x+4)^2+(y)^2+(z)^2}=25+4x$

Now again squaring both sides,

$25\left ( {(x+4)^2+(y)^2+(z)^2} \right )=625+200x+16x^2$

$25x^2+200x+400+25y^2+25z^2=625+200x+16x^2$

$9x^2+25y^2+25z^2-225=0$

Hence the equation of the set of points P, the sum of whose distances from A and B is equal to 10 is $9x^2+25y^2+25z^2-225=0$ .

Class 11 maths chapter 12 NCERT solutions - Exercise: 12.3

Question:1(i) Find the coordinates of the point which divides the line segment joining the points (– 2, 3, 5) and (1, – 4, 6) in the ratio 2 : 3 internally

Answer:

The line segment joining the points A (– 2, 3, 5) and B(1, – 4, 6)

Let point P(x,y,z) be the point that divides the line segment AB internally in the ratio 2:3.

Now, As we know by section formula, The coordinate of the point P which divides line segment $A(x_1,y_1,z_1)$ And $B(x_2,y_2,z_2)$ in ratio m:n is

$\left (\frac{mx_2+nx_1}{m+n},\frac{my_2+ny_1}{m+n},\frac{mz_2+nz_1}{m+n} \right )$

Now the point that divides A (– 2, 3, 5) and B(1, – 4, 6) in ratio 2:3 is

$\left (\frac{2(1)+3(-2)}{2+3},\frac{2(-4)+3(3)}{2+3},\frac{2(6)+3(5)}{2+3} \right )=\left ( \frac{-4}{5},\frac{1}{5},\frac{27}{5} \right )$

Hence required point is

$\left ( \frac{-4}{5},\frac{1}{5},\frac{27}{5} \right )$ .

Question:1 (ii) Find the coordinates of the point which divides the line segment joining the points (– 2, 3, 5) and (1, – 4, 6) in the ratio 2 : 3 externally.

Answer:

he line segment joining the points A (– 2, 3, 5) and B(1, – 4, 6)

Let point P(x,y,z) be the point that divides the line segment AB externally in the ratio 2:3.

Now, As we know by section formula , The coordinate of the point P which divides line segment $A(x_1,y_1,z_1)$ And $B(x_2,y_2,z_2)$ externally in ratio m:n is

$\left (\frac{mx_2-nx_1}{m-n},\frac{my_2-ny_1}{m-n},\frac{mz_2-nz_1}{m-n} \right )$

Now the point that divides A (– 2, 3, 5) and B(1, – 4, 6) externally in ratio 2:3 is

$\left (\frac{2(1)-3(-2)}{2-3},\frac{2(-4)-3(3)}{2-3},\frac{2(6)-3(5)}{2-3} \right )=\left ( -8,17,3\right )$

Hence required point is

. $( -8,17,3)$

Question:2 Given that P (3, 2, – 4), Q (5, 4, – 6) and R (9, 8, –10) are collinear. Find the ratio in which Q divides PR.

Answer:

Given Three points,

P (3, 2, – 4), Q (5, 4, – 6) and R (9, 8, –10)

Let point Q divides PR internally in the ratio $\lambda:1$

Now,

According to the section formula , The point Q in terms of P,Q and $\lambda$ is:

$\left ( \frac{9\lambda+3}{\lambda+1},\frac{8\lambda+2 }{\lambda +1},\frac{-10\lambda-4}{\lambda+1}\right )=(5,4,-6)$

$\frac{9\lambda+3}{\lambda+1}=5$

$4\lambda =2$

$\lambda =\frac{1}{2}$

Hence, point Q divides PR in ratio 1:2.

Question:3 Find the ratio in which the YZ-plane divides the line segment formed by joining the points (–2, 4, 7) and (3, –5, 8).

Answer:

Given,

two points A(–2, 4, 7) and B(3, –5, 8)

Let Y-Z plane divides AB in $\lambda:1$

So, According to the section formula, the point which divides AB in $\lambda:1$ is

$\left ( \frac{3\lambda-2}{\lambda+1},\frac{-5\lambda +4}{\lambda+1},\frac{8\lambda +7}{\lambda+1} \right )$

Since this point is in YZ plane, x coordinate of this point will be zero.

So,

$\frac{3\lambda-2}{\lambda+1}=0$

${3\lambda-2}=0$

$\lambda = \frac{2}{3}$

Hence YZ plane divides Line segment AB in a ratio 2:3.

Question:4 Using section formula, show that the points A (2, –3, 4), B (–1, 2, 1) and $C \left ( 0 , 1/3 , 2 \right )$ are collinear.

Answer:

Given,

three points, A (2, –3, 4), B (–1, 2, 1) and C (0, 1/3, 2)

Let a point P divides Line segment AB in the ratio $\lambda:1$

SO, according to the section formula, the point P will be

$\left (\frac{-\lambda +2}{\lambda+1},\frac{2\lambda-3}{\lambda+1},\frac{\lambda+4}{\lambda+1} \right )$

Now, let's compare this point P with point C.

$\frac{-\lambda +2}{\lambda+1},=0$

$\lambda=2$

From here, we see that for $\lambda=2$ , point C divides the line segment AB in ratio 2:1. Since point C divides the line segment AB, it lies in the line joining A and B and Hence they are colinear.

Question:5 Find the coordinates of the points which trisect the line segment joining the points P (4, 2, – 6) and Q (10, –16, 6).

Answer:

Given,

two points P (4, 2, – 6) and Q (10, –16, 6).

The point which trisects the line segment are the points which divide PQ in either 1:2 or 2:1

Let R (x,y,z) be the point which divides Line segment PR in ratio 1:2

Now, according to the section formula

$(x,y,z)=\left ( \frac{10+2(4)}{1+2},\frac{-16+2(2)}{1+2},\frac{6-2(6)}{1+2} \right )=(6,-4,-2)$

Let S be the point which divides the Line segment PQ in ratio 2:1

So, The point S according to section formula is

$(x,y,z)=\left ( \frac{(2)10+(4)}{1+2},\frac{2(-16)+(2)}{1+2},\frac{(2)6-(6)}{1+2} \right )=(8,-10,2)$

Hence the points which trisect the line segment AB are (6,-4,-2) and (8,-10,2).

Introduction to three dimensional geometry NCERT solutions - Miscellaneous Exercise

Question:1 Three vertices of a parallelogram ABCD are A(3, – 1, 2), B (1, 2, – 4) and C (– 1, 1, 2). Find the coordinates of the fourth vertex.

Answer:

Given

Three vertices of the parallelogram, ABCD are:

A(3, – 1, 2), B (1, 2, – 4) and C (– 1, 1, 2).

Let the fourth vertice be (a, b, c )

Now, As we know the concept that diagonal of the parallelogram bisect each other,

Hence here in parallelogram ABCD, the midpoint of the line segment AC will be equal to the midpoint of the line segment BD.

So,

$\left ( \frac{3-1}{2},\frac{-1+1}{2},\frac{2+2}{2} \right )=\left ( \frac{1+a}{2},\frac{2+b}{2},\frac{-4+c}{2} \right )$

On comparing both points, we get

$a=1,b=-2\:and\:c=8$

Hence the Fourth vertice of the is (1,-2,8)

Question:2 Find the lengths of the medians of the triangle with vertices A (0, 0, 6), B (0,4, 0) and (6, 0, 0).

Answer:

Given,

Three vertices of the triangle,A (0, 0, 6), B (0,4, 0)and C(6, 0, 0).

Now,

Let D be the midpoint of the AB, E be the midpoint of the BC and F be the midpoint of the AC.

Vertice of the D =

$\left ( \frac{0+0}{2},\frac{0+4}{2},\frac{6+0}{2} \right )=(0,2,3)$

Vertices of E =

$\left ( \frac{0+6}{2},\frac{4+0}{2},\frac{0+0}{2} \right )=(3,2,0)$

Vertices of the F =

$\left ( \frac{0+6}{2},\frac{0+0}{2},\frac{6+0}{2} \right )=(3,0,3)$

Now, Medians of the triangle are CD, AE, and BF

So the lengths of the medians are:

$CD=\sqrt{}$ $CD=\sqrt{(6-0)^2+(0-2)^2+(0-3)^2}=\sqrt{36+4+9}=\sqrt{49}=7$

$AE=\sqrt{(0-3)^2+(0-2)^2+(6-0)^2}=\sqrt{9+4+36}=\sqrt{49}=7$

$BF=\sqrt{(3-0)^2+(0-4)^2+(3-0)^2}=\sqrt{9+16+9}=\sqrt{34}$

Hence lengths of the median are $7,7 \:and\:\sqrt{34}$ .

Question:3 If the origin is the centroid of the triangle PQR with vertices P (2a, 2, 6), Q (– 4, 3b, –10) and R(8, 14, 2c), then find the values of a, b and c.

Answer:

Given,

Triangle PQR with vertices P (2a, 2, 6), Q (– 4, 3b, –10) and R(8, 14, 2c),

Now, As we know,

The centroid of a triangle is given by

$\left (\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3} ,\frac{z_1 +z_2+z_3}{3}\right )$

Where coordinates of vertices of the triangle are $(x_1,x_2,x_3),(y_1,y_2,y_3)\:and\:(z_1,z_2,z_3)$

Since Centroid of the triangle, PQR is origin =(0,0,0)

$\left (\frac{2a-4+8}{3},\frac{2+3b+14}{3},\frac{6-10+2c}{3} \right )=(0,0,0)$

On equating both coordinates, we get

$a=-2,b=\frac{-16}{3}\:and\:c=2$

Question:4 Find the coordinates of a point on y-axis which are at a distance of $5 \sqrt 2$ from the point P (3, –2, 5).

Answer:

Let the point Q be (0,y,0)

Now Given

The distance of point Q From point P = $5 \sqrt 2$

So,

$\sqrt{(3-0)^2+(-2-y)^2+(5-0)^2}=5\sqrt{2}$

${(3-0)^2+(-2-y)^2+(5-0)^2}=25\times 2$

$9+(2+y)^2+25=50$

$(2+y)^2=16$

$(2+y)=4\:or\:2+y=-4$

$y=2\:or\:y=-6$

The coordinates of the required point is (0,2,0) or (0,-6,0).

Question:5 A point R with x-coordinate 4 lies on the line segment joining the points P(2, –3, 4) and Q (8, 0, 10). Find the coordinates of the point R. [Hint Suppose R divides PQ in the ratio k : 1. The coordinates of the point R are given by $\left ( \frac{8k + 2 }{k+1}, \frac{-3}{k+1}, \frac{10 k + 4 }{k+1} \right )$

Answer:

Let R divides PQ in the ratio k: 1.

The coordinates of the point R are given by

$\left ( \frac{8k + 2 }{k+1}, \frac{-3}{k+1}, \frac{10 k + 4 }{k+1} \right )$

now, as given the x coordinate of the R is 4.

So,

$\frac{8k+2}{k+1}=4$

$k=\frac{1}{2}$

Hence the coordinates of R are:

$\left ( \frac{8\times\frac{1}{2} + 2 }{\frac{1}{2}+1}, \frac{-3}{\frac{1}{2}+1}, \frac{10 \times \frac{1}{2}+ 4 }{\frac{1}{2}+1} \right )=(4,-2,6)$

Question:6 If A and B be the points (3, 4, 5) and (–1, 3, –7), respectively, find the equation of the set of points P such that $PA ^2 + PB ^ 2 = k^ 2$ , where k is a constant.

Answer:

Given Points,

A (3, 4, 5) and B (–1, 3, –7),

Let the coordinates of point P be (x,y,z)

Now,

Given condition :

$PB^2=(x-(-1))^2+(y-(3))^2+(z-(-7))^2$

$PB^2=x^2+2x+1+y^2-6y+9+z^2+14z+49$

$PB^2=x^2+y^2+z^2+2x-6y+14z+59$

And

$PA^2=(x-3)^2+(y-4)^2+(z-5)^2$

$PA^2=x^2+y^2+z^2-6x-8y-10z+50$

Now, Given Condition

$PA ^2 + PB ^ 2 = k^ 2$

$x^2+y^2+z^2+2x-6y+14z+59+x^2+y^2+z^2-6x-8y-10z+50=k^2$

$2x^2+2y^2+2z^2-4x-14y+4z+109=k^2$

Hence Equation of the set of the point P is

$2x^2+2y^2+2z^2-4x-14y+4z+109=k^2$ .

Summary of Three Dimensional Geometry Class 11 Maths

12.1 Introduction

12.2 Coordinate Axes and Coordinate Planes in Three Dimensional Space

12.3 Coordinates of a Point in Space

12.4 Distance between Two Points

12.5 Section Formula

Interested students can practice class 11 maths ch 12 question answer using the given exercises.

Unlike 2D shapes, 3D shapes have length, width and depth or thickness. In three dimensional geometry, a coordinate space is determined by a vector perpendicular to that particular plane. Thus, a point in a three-dimensional plane will include the coordinates of three planes named as X- plane, Y- plane, and Z-plane.

There are some important formulas from NCERT solutions for class 11 maths chapter 12 introduction to three dimensional geometry which you should remember after studying this chapter.

Distance between two points $P(x_{1}, y_{1}, z_{1})\hspace{0.5cm}and \hspace{0.5cm} Q(x_{2}, y_{2}, z_{2})$ is

$\mathrm{PQ}=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}+\left(z_{2}-z_{1}\right)^{2}}$

The coordinates of the point R which divides the line segment joining two points $P(x_{1}, y_{1}, z_{1})\hspace{0.5cm} and \hspace{0.5cm} Q(x_{2}, y_{2}, z_{2})$ internally in the ratio m : n

$\left(\frac{m x_{2}+n x_{1}}{m+n}, \frac{m y_{2}+n y_{1}}{m+n}, \frac{m z_{2}+n z_{1}}{m+n}\right)$

The coordinates of the point R which divides the line segment joining two points $P(x_{1}, y_{1}, z_{1})\hspace{0.5cm} and \hspace{0.5cm} Q(x_{2}, y_{2}, z_{2})$ externally in the ratio m : n

$\left(\frac{m x_{2}-n x_{1}}{m-n}, \frac{m y_{2}-n y_{1}}{m-n}, \frac{m z_{2}-n z_{1}}{m-n}\right)$

Distance between origin $P(0,0,0)\hspace{0.5cm} and \hspace{0.5cm} Q(x_{2}, y_{2}, z_{2})$ is

$\mathrm{OQ}=\sqrt{x_{2}^{2}+y_{2}^{2}+z_{2}^{2}}$

The coordinates of the mid-point of the line segment joining two points $P(x_{1}, y_{1}, z_{1})\hspace{0.5cm} and \hspace{0.5cm} Q(x_{2}, y_{2}, z_{2})$ is

$\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}, \frac{z_{1}+z_{2}}{2}\right)$

The coordinates of the centroid of the triangle, whose vertices are $\left (x_{1}, y_{1}, z_{1}\right), \left (x_{2}, y_{2}, z_{2}\right),\left (x_{3}, y_{3}, z_{3}\right)$

$\left(\frac{x_{1}+x_{2}+x_{3}}{3}, \frac{y_{1}+y_{2}+y_{3}}{3}, \frac{z_{1}+z_{2}+x_{3}}{3}\right)$

NCERT Solutions for Class 11 Mathematics

chapter-1	Sets
chapter-2	Relations and Functions
chapter-3	Trigonometric Functions
chapter-4	Principle of Mathematical Induction
chapter-5	Complex Numbers and Quadratic equations
chapter-6	Linear Inequalities
chapter-7	Permutation and Combinations
chapter-8	Binomial Theorem
chapter-9	Sequences and Series
chapter-10	Straight Lines
chapter-11	Conic Section
chapter-12	Introduction to Three Dimensional Geometry
chapter-13	Limits and Derivatives
chapter-14	Mathematical Reasoning
chapter-15	Statistics
chapter-16	Probability

Key Features Of Introduction to Three Dimensional Geometry Class 11 Solutions

Step-by-step solution: To understand the concepts of 3d geometry class 11, it is important to solve problems step-by-step, which helps students to grasp the concepts and develop problem-solving skills.

Comprehensive coverage: The chapter 12 maths class 11 is covered in a comprehensive manner in Class 11 Mathematics, with a focus on understanding the fundamental concepts and their applications in solving problems.

Concepts: The topic of introduction to coordinate geometry class 11 involves the study of various concepts such as co-ordinate geometry, lines and planes in space, distance formula, and direction ratios.

NCERT Solutions for Class 11 - Subject wise

NCERT Books and NCERT Syllabus

Happy Reading !!!

Frequently Asked Questions (FAQs)

1. Explain the concept of 3D Geometry covered in class 11 maths chapter 12.

Three Dimensional Geometry ch 12 maths class 11 is a field of Mathematics that focuses on the analysis of points, lines, and solid shapes in three-dimensional coordinate systems. Introduction to 3d geometry class 11 introduces students to the concept of the z-coordinate in addition to the x and y coordinates, which is used to determine the precise location of a point in the 3D coordinate plane. It is a fundamental theory with broad applications in other scientific disciplines and advanced Mathematics. Additionally, the trigonometric ratios concept has important applications in 3D Geometry.

2. Discuss the topics covered in chapter 12 class 11 maths.

maths chapter 12 class 11 includes the following topics:

Introduction
Coordinate Axes and Coordinate Planes in Three Dimensional Space
Point Coordinates in 3D Space
Distance Formula between Two Points
Section Formula

3. Where can I find the complete solutions of NCERT for class 11 maths ?

Students can find detailed NCERT solutions for class 11 maths by clicking on the link. they can practice problems given in introduction to three dimensional geometry class 11 to get good hold on the concepts. that ultimately benefit during the exam and lead to good score.

4. How many chapters are there in CBSE class 11 maths ?

There are 16 chapters starting from sets to probability in the CBSE class 11 maths.

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NCERT Solutions for Miscellaneous Exercise Chapter 16 Class 11 - Probability

NCERT Exemplar Class 11 Maths Solutions Chapter 12 Introduction to Three Dimensional Geometry

NCERT Exemplar Class 11 Maths Solutions Chapter 10 Straight Lines

NCERT Exemplar Class 11 Maths Solutions Chapter 6 Linear Inequalities

NCERT Exemplar Class 11 Maths Solutions Chapter 16 Probability

NCERT Exemplar Class 11 Maths Solutions Chapter 14 Mathematical Reasoning

NCERT Exemplar Class 11 Maths Solutions Chapter 7 Permutations and Combinations

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

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Option 3) Fraction of solute present in water	Option 4) Mole fraction.

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NCERT Solutions for Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry

Introduction to Three Dimensional Geometry Class 11 Questions And Answers

Introduction to Three Dimensional Geometry Class 11 Questions And Answers PDF Free Download

Introduction to Three Dimensional Geometry Class 11 Solutions - Important Formulae

Introduction to Three Dimensional Geometry Class 11 NCERT Solutions (Intext Questions and Exercise)

Summary of Three Dimensional Geometry Class 11 Maths

NCERT Solutions for Class 11 Mathematics

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