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

Edited By Ravindra Pindel | Updated on Aug 22, 2022 - 12:45 p.m. IST

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 NCERT solutions 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 NCERT syllabus chapter is new for you but you can easily understand the concepts with the help of NCERT solutions for class 11 maths chapter 12 introduction to three dimensional geometry.

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It is important to pay careful attention and try to relate the 2D geometry concepts with this chapter while studying this NCERT book chapter. This chapter is very important for CBSE class 11 final examination and in various competitive exams like BITSAT, JEE Main, etc. In introduction to three Dimensional geometry class 11 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. Here you will get NCERT solutions for class 11 also.

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.

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

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:

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

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

$4\lambda =2$

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

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-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

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

Download NCERT Solutions For Class 11 Maths Chapter 12 Exercise Wise-

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 Class 11 Maths Chapter 1 2 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

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

NCERT Solutions for Class 11- Subject wise

Also Check NCERT Books and NCERT Syllabus here

Introduction to Three Dimensional Geometry Class 11 Maths Chapter Points To Remember

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

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

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 NCERT solutions for class 11 maths chapter 12 introduction to three dimensional geometry, you will get solutions of miscellaneous exercise too.

Happy Reading !!!

Solutions

Frequently Asked Question (FAQs) - NCERT Solutions for Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry

Question: Which chapters are the most difficult chapters in the CBSE class 11 maths ?

Answer:

Permutations and combinations, trigonometric functions are considered to be the most difficult chapters in the CBSE class 11 maths but the rigorous practice you will get command on them also.

Question: How many chapters are there in CBSE class 11 maths ?

Answer:

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

Question: Where can I find the complete solutions of NCERT for class 11 maths ?

Answer:

Here you will get the detailed NCERT solutions for class 11 maths by clicking on the link.

Question: Does CBSE provides the solutions of NCERT class 11 maths ?

Answer:

No, CBSE doesn’t provide NCERT solutions for any class or subject.

Question: What is the weightage of the chapter sequence and series in Jee Main ?

Answer:

This chapter has about 4% weightage in the Jee main exam.

Question: What are the important topics of the chapter Introduction to Three Dimensional Geometry ?

Answer:

Coordinate axes and coordinate planes in three-dimensional space, coordinates of a point in space, the distance between two points, and section formula are the important topics of this chapter.

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

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NCERT Solutions

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

Introduction To Three Dimensional Geometry-Exercise: 12.1

Introduction To Three Dimensional Geometry-Exercise: 12.2

Introduction To Three Dimensional Geometry-Exercise: 12.3

Introduction To Three Dimensional Geometry-Miscellaneous Exercise

Download NCERT Solutions For Class 11 Maths Chapter 12 Exercise Wise-

NCERT Solutions for Class 11 Mathematics

NCERT Solutions for Class 11- Subject wise

Also Check NCERT Books and NCERT Syllabus here

Introduction to Three Dimensional Geometry Class 11 Maths Chapter Points To Remember

Frequently Asked Question (FAQs) - NCERT Solutions for Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry

Question: Which chapters are the most difficult chapters in the CBSE class 11 maths ?

Question: How many chapters are there in CBSE class 11 maths ?

Question: Where can I find the complete solutions of NCERT for class 11 maths ?

Question: Does CBSE provides the solutions of NCERT class 11 maths ?

Question: What is the weightage of the chapter sequence and series in Jee Main ?

Question: What are the important topics of the chapter Introduction to Three Dimensional Geometry ?