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

Class 11 Maths Chapter 11 Question Answer
Exercise: 11.1, Page: 211, Total Questions: 4

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

Answer:

Any point on x-axis has 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,e 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 11 question answer -
Exercise: 11.2, Page: 213, Total questions: 5

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

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

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

$=\sqrt{20}$

$=2\sqrt{5}$

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

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

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

$=\sqrt{104}$

$=2\sqrt{26}$

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

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

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

$=\sqrt{20}$

$=2\sqrt{5}$

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

Answer:

Given,three 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 :

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

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

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

$=\sqrt{14}$

The distance BC:

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

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

$=\sqrt{56}$

$=2\sqrt{14}$

The distance CA

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

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

$=\sqrt{126}$

$=3\sqrt{14}$

As we can see here,

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

Hence, we can say that points 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

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

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

$=\sqrt{18}$

The distance BC

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

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

$=\sqrt{18}$

The distance CA

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

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

$=\sqrt{36}$

$=6$

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

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

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

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

$=\sqrt{18}$

The distance BC

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

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

$=\sqrt{18}$

The distance CA

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

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

$=\sqrt{36}$

$=6$

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

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

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

$=\sqrt{36}$

$=6$

The distance BC

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

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

$=\sqrt{43}$

The distance CD

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

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

$=\sqrt{36}$

$=6$

The distance DA

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

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

$=\sqrt{43}$

Here, As we can see

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

As the opposite sides of the 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

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

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

⇒ $[(x-1{)}^2-(x-3{)}^2]+[(y-2{)}^2-(y-2{)}^2]+[(z-3{)}^2-(z+1{)}^2]=0$

Now lets apply the simplification property,

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

⇒ $[(2)(2x-4)]+0+[(-4)(2z-2)]=0$

⇒ $4x-8-8z+8=0$

⇒ $4x-8z=0$

⇒ $x-2z=0$

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 11 question answer -
Exercise: Miscellaneous, Page: 215, Total questions: 4

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

On comparing both points, we get

$a=1,b=-2andc=8$

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 =

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

Vertices of E =

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

Vertices of the F =

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

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 $7,7and\sqrt{34}$ .

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

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

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

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

On equating both coordinates, we get

$a=-2,b=\frac{-16}{3}andc=2$

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,

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

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

⇒ $9+(2+y{)}^2+25=50$

⇒ $(2+y{)}^2=16$

⇒ $(2+y)=4or2+y=-4$

⇒ $y=2ory=-6$

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

Answer:

Let R divides PQ in the ratio k: 1.

The coordinates of the point R are given by

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

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:

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

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 $P{A}^2+P{B}^2=k^2$ , 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 :

$P{B}^2=(x-(-1){)}^2+(y-(3){)}^2+(z-(-7){)}^2$

⇒ $P{B}^2=x^2+2x+1+y^2-6y+9+z^2+14z+49$

⇒ $P{B}^2=x^2+y^2+z^2+2x-6y+14z+59$

And

$P{A}^2=(x-3{)}^2+(y-4{)}^2+(z-5{)}^2$

⇒ $P{A}^2=x^2+y^2+z^2-6x-8y-10z+50$

Now, Given Condition

$P{A}^2+P{B}^2=k^2$

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

⇒ $2x^2+2y^2+2z^2-4x-14y+4z+109=k^2$

Hence Equation of the set of the point P is

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