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

 

NCERT solutions for class 11 maths chapter 12 Introduction to Three Dimensional Geometry: In our previous classes, you have studied the basic concepts of geometry in two-dimensional space. In this article, you will get solutions of NCERT for class 11 maths chapter 12 introduction to three dimensional geometry. A 3D shape can be defined as an object or a solid figure or shape that has 3 dimensions in length, width, and height. This chapter is new for you but you can easily understand the concepts with the help of CBSE NCERT solutions for class 11 maths chapter 12 introduction to three dimensional geometry. It is important to pay careful attention and try to relate the 2D geometry concepts with this chapter while studying this chapter. This chapter is very important for CBSE class 11 final examination and in various competitive exams like BITSAT, JEE Mains, etc. In this chapter, you will learn what is 3D 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 it on your own. If you are not able to do, you can take the help of NCERT solutions for class 11 maths chapter 12 introduction to three dimensional geometry which is prepared in a detailed manner. Check all NCERT solutions from class 6 to class 12 to learn CBSE maths.

What is the 3D Coordinate Geometry?

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.                                        

Topics of NCERT grade 11 maths chapter 12 Introduction to Three Dimensional Geometry

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

The complete solutions of NCERT class 11 mathematics chapter 12 is provided below:

NCERT solutions for class 11 maths chapter 12 introduction to three dimensional geometry-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.

Solutions of NCERT for class 11 maths chapter 12 introduction to three dimensional geometry-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.

CBSE NCERT solutions for class 11 maths chapter 12 introduction to three dimensional geometry-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

{9\lambda+3}}=5(\lambda+1)

{9\lambda+3}}=5\lambda+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). 

NCERT solutions for class 11 maths chapter 12 introduction to three dimensional geometry-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.

NCERT solutions for class 11 mathematics

chapter-1

NCERT solutions for class 11 maths chapter 1 Sets

chapter-2

Solutions of NCERT for class 11 chapter 2 Relations and Functions

chapter-3

CBSE NCERT solutions for class 11 chapter 3 Trigonometric Functions

chapter-4

NCERT solutions for class 11 chapter 4 Principle of Mathematical Induction

chapter-5

Solutions of NCERT for class 11 chapter 5 Complex Numbers and Quadratic equations

chapter-6

CBSE NCERT solutions for class 11 maths chapter 6 Linear Inequalities

chapter-7

NCERT solutions for class 11 maths chapter 7 Permutation and Combinations

chapter-8

Solutions of NCERT for class 11 maths chapter 8 Binomial Theorem

chapter-9

CBSE NCERT solutions for class 11 maths chapter 9 Sequences and Series

chapter-10

NCERT solutions for class 11 maths chapter 10 Straight Lines

chapter-11

Solutions of NCERT for class 11 maths chapter 11 Conic Section

chapter-12

CBSE NCERT solutions for class 11 maths chapter 12 Introduction to Three Dimensional Geometry

chapter-13

NCERT solutions for class 11 maths chapter 13 Limits and Derivatives

chapter-14

Solutions of NCERT for class 11 maths chapter 14 Mathematical Reasoning

chapter-15

CBSE NCERT solutions for class 11 maths chapter 15 Statistics

chapter-16

NCERT solutions for class 11 maths chapter 16 Probability

NCERT solutions for class 11- Subject wise

Solutions of NCERT for class 11 biology

CBSE NCERT solutions for class 11 maths

NCERT solutions for class 11 chemistry

Solutions of NCERT for Class 11 physics

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 \left P(x_{1}, y_{1}, z_{1}\right) \text { and }\left Q(x_{2}, y_{2}, z_{2}\right) 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 \left P(x_{1}, y_{1}, z_{1}\right) \text { and }\left Q(x_{2}, y_{2}, z_{2}\right) 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 \left P(x_{1}, y_{1}, z_{1}\right) \text { and }\left Q(x_{2}, y_{2}, z_{2}\right) 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 \left O(0,0,0\right) \text { and }\left Q(x_{2}, y_{2}, z_{2}\right) 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 \left P(x_{1}, y_{1}, z_{1}\right) \text { and }\left Q(x_{2}, y_{2}, z_{2}\right) 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)

Tip- All the  3D geometry formulas are very similar to 2D geometry, so try to relate it with 2D geometry it will very easy for you to remember the formulas. You should write the formulas which you are using in solving problems. There are 6 exercises in the miscellaneous exercise. You should solve that also to get command on this chapter.  In CBSE NCERT solutions for class 11 maths chapter 12 introduction to three dimensional geometry, you will get solutions of miscellaneous exercise too.  

Happy Reading !!! 

 

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