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

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

Edited By Ravindra Pindel | Updated on Sep 12, 2022 06:06 PM IST

NCERT Exemplar Class 11 Maths Solutions Chapter 12 are provided to the students after thorough study and research of three-dimensional geometry, and different methods and techniques of problem-solving by experts to provide very easy yet convenient and understanding solutions to help and provide guidance for exams. NCERT Class 11 Maths exemplar solutions chapter 12 deals with concepts relating to three-dimensional geometry with the progression of coordinate geometry in the three-dimensional plane to the two-dimensional plane from the last two chapters along with its application in different methods. NCERT Exemplar Class 11 Maths chapter 12 solutions are detailed explanations for the problems of NCERT created through the careful and professional guidance of experts for the preparation and examination of students from the perspective of Board exams.
Also, check - NCERT Class 11 Maths Solutions.

This Story also Contains
  1. NCERT Class 11 Exemplar Maths Solutions Chapter 12 - Question-Wise Solution
  2. More about NCERT Exemplar Class 11 Maths Solutions Chapter 12
  3. NCERT Exemplar Class 11 Maths Solutions Chapter-Wise
  4. Important Topics To Cover From NCERT Exemplar Solutions For Class 11 Maths Chapter 12

NCERT Class 11 Exemplar Maths Solutions Chapter 12 - Question-Wise Solution

Question:1

Locate the following points:
(i) (1, – 1, 3), (ii) (– 1, 2, 4)
(iii) (– 2, – 4, –7) (iv) (– 4, 2, – 5).

Answer:

Question:2

Name the octant in which each of the following points lies.
(i) (1, 2, 3), (ii) (4, – 2, 3), (iii) (4, –2, –5) (iv) (4, 2, –5)
(v) (– 4, 2, 5) (vi) (–3, –1, 6) (vii) (2, – 4, – 7) (viii) (– 4, 2, – 5).

Answer:

The octant in which a point lies is determined by the signs of the coordinates of the point.

Octants

Coordinates

I

II

III

IV

V

VI

VII

VIII

x

+

-

-

+

+

-

-

+

y

+

+

-

-

+

+

-

-

Z

+

+

+

+

-

-

-

-

Thus,

  1. (1,2,3) → first quadrant

  2. (4,-2,3) → Fourth octant

  3. (4,-2,-5) → eighth octant

  4. (4,2,-5) → fifth octant

  5. (-4,2,5) → second octant

  6. (-3,-1,6) → third octant

  7. (3,-4,-7) → eighth octant

  8. (-4,2,-5) → sixth octant

Question:3

Let A, B, C be the feet of perpendiculars from a point P on the x, y, z-axis respectively. Find the coordinates of A, B and C in each of the following where the point P is :
(i) A = (3, 4, 2) (ii) (–5, 3, 7) (iii) (4, – 3, – 5)

Answer:

We already know that,
On x-axis, y,z= 0
On y-axis, x,z= 0
On z-axis, x,y = 0
Thus, from point P, the feet of perpendiculars will be-

  1. A(3,0,0), 5(0,4,0), C(0,0,2)

  2. A(-5,0,0), B(0,3,0), C(0,0,7)

  3. A(4,0,0), 5(0,-3,0), C(0,0,-5)

Question:4

Let A, B, C be the feet of perpendiculars from a point P on the xy, yz and zxplanes respectively. Find the coordinates of A, B, C in each of the following where the point P is
(i) (3, 4, 5) (ii) (–5, 3, 7) (iii) (4, – 3, – 5).

Answer:

We know that,
On xy-plane, z = 0
On yz-plane, x=0
On zx-plane, y = 0
Thus, the coordinates from the perpendicular to different planes from the given points are,

  1. A(3,4,0), 5(0,4,5), C(3,0,5)

  2. A(-5,3,0), 5(0,3,7), C(-5,0,7)

  3. A(4,-3,0), 5(0,-3,-5), C(4,0,-5)

Question:5

How far apart are the points (2, 0, 0) and (–3, 0, 0)?

Answer:

AB =| 2-(-3) |
= 5

Question:6

Find the distance from the origin to (6, 6, 7).

Answer:

From Origin to Point (6,6,7),
Distance =\sqrt{\left ( 0-6 \right )^{2}+\left ( 0-6 \right )^{2}+\left ( 0-7 \right )^{2}}
= \sqrt{121}
= 11

Question:7

Show that if x square plus y square equal to 1, then the point x y square root of 1 minus x square minus y square is at a distance 1 unit from the origin.
Answer:

Given: x2 + y2 = 1
From origin to point \left ( x,y ,\sqrt{1 - x^{2}-y^{2}} \right ) )
Distance, d= \sqrt{x^{2}+y^{2}+ \left (\sqrt{1 - x^{2}-y^{2}} \right )^{2} }
= \sqrt{x^{2}+y^{2}+1 - x^{2}-y^{2}}=1

Question:8

Show that the point A (1, – 1, 3), B (2, – 4, 5) and (5, – 13, 11) are collinear.

Answer:

Given points:
4(1,-1,3), 6(2,-4,5), C(5,-13,11)
Thus,
AB=\sqrt{(1-2)^{2}+(-1+4)^{2}+(3-5)^{2}}
=\sqrt{1+9+4}=\sqrt{14}
BC=\sqrt{(2-5)^{2}+(-4+13)^{2}+(5-11)^{2}}=3\sqrt{14}
AC=\sqrt{(1-5)^{2}+(-1+13)^{2}+(3-11)^{2}}=4\sqrt{14}
Now,
AB + BC = =\sqrt{14}+3\sqrt{14}
= 4\sqrt{14} = AC

Question:9

Three consecutive vertices of a parallelogram ABCD are A (6, – 2, 4), B (2, 4, – 8), C (–2, 2, 4). Find the coordinates of the fourth vertex

Answer:

Let us assume that the coordinates of the fourth vertex D is (x,y,z).
Now, mid-point of diagonal AC = P(6-2/2, -2+2/2,4+4/2)
= P(2,0,4)
Since it is a parallelogram, the mid-point of BD will also be P,
= P(x+2/2,y+4/2,z-8/2)
= P(2,0,4)
Now, let us equate the coordinates of P,
X+2/2 = 2, i.e., x = 2;
Y+4/2 = 0, i.e., y = -4;
Z-8/2 = 4, i.e., z = 16
Therefore, the coordinates of D will be (2,-4,16).

Question:10

Show that the triangle ABC with vertices A (0, 4, 1), B (2, 3, – 1) and C (4, 5, 0) is right angled

Answer:

Vertices of ?ABC are –
A(0,4,1), 5(2,3,-1), C(4,5,0)
Thus,
AB = \sqrt{(0-2)^{2}+(4-3)^{2}+(1+1)^{2} }
= \sqrt{4+1+1 } = 3
BC = \sqrt{(2-4)^{2}+(3-5)^{2}+(-1-0)^{2} }
= \sqrt{4+4+1 } = 3
AC = \sqrt{(0-4)^{2}+(4-5)^{2}+(1-0)^{2} }
= \sqrt{16+1+1 } = \sqrt{18}
Thus, it is clear that,
AC2 = AB2 + BC2,
Hence it is a right angled triangle.

Question:11

Find the third vertex of triangle whose centroid is origin and two vertices are (2, 4, 6) and (0, –2, –5).

Answer:

Given:
Vertices of triangle,
B(2, 4, 6), C(0,-2,-5)
& centroid of triangle, G is (0,0,0)
Let us assume that the unknown vertex of the triangle is A(x,y,z).
Thus,
(0,0,0) ≡ ((2+0+x)/3, (4-2+y)/3, (6-5+z)/3)
Now, when we compare the coordinates, we get,
2+x/3 = 0, thus, x = -2
2+y/3 = 0, thus, y = -2
& 1+z/3 = 0, thus, z = -1.

Question:12

Find the centroid of a triangle, the mid-point of whose sides are D (1, 2, – 3), E (3, 0, 1) and F (– 1, 1, – 4).

Answer:

Given:
Mid-points of sides of triangle ABC are,
D(1,2,-3), E(3,0,1) & F(-1,1,4)
Now, the centroid of triangle DEF = Centroid of triangle ABC ……. (from geometry of centroid)
Thus, centroid of triangle ABC,
G(1+3-1/3, 2+0+1/3, -3+1-4/3) ≡ G(1,1,-2)

Question:13

The mid-points of the sides of a triangle are (5, 7, 11), (0, 8, 5) and (2, 3, – 1). Find its vertices.

Answer:

Given:
Mid-points of sides of triangle ABC are,
D(5,7,11), E(0,8,5) & F(2,3,-1)
Let us assume that the vertices of the triangle are –
A(x1, y1, z1), B(x2, y2, z2) & C(x3, y3, z3).
Now, we know that E is the mid-point of AC, thus,
(x1 + x3/2, y1 + y3/2, z1 + z3/2) = (0,8,5)
Thus,
C(x3, y3, z3) = C(-x1, 16 – y1, -2-z­1)
Now, mid-point of AB, F is
(x1 + x2/2, y1 + y2/2, z1 + z2/2) = (2,3,-1)
Now, B(x2, y2, z2) = B(4-x1, 6-y1, -1-z1)
Now, mid-point of BC, D is,
-x1 + 4 – x1/2 = 5,
16 – y1 + 6 – y1/2 = 7
& 10 – z1 – 2 – z1/2 = 11
Thus, x1 = -3, y1 = 4 & z1 = -7
Thus, A = (-3,4,-7), B = (7,2,5) & C = (3,12,17)

Question:14

Three vertices of a Parallelogram ABCD are A (1, 2, 3), B (– 1, – 2, – 1) and C (2, 3, 2). Find the fourth vertex D.

Answer:

Let us assume that the 4th vertex if the parallelogram is D(x,y,z).
Now, mid-point of BD,
(1+2/2, 2+3/2, 3+2/2) = (x-1/2, y-2/2, z-1/2)
Thus, 3/2 = x-1/2
x = 4
5/2 = y-2/2
y = 7
& 5/2 = z-1/2
Thus, z = 6.
Therefore, coordinates of the 4th vertex are D(4,7,6).

Question:15

Find the coordinate of the points which trisect the line segment joining the points A (2, 1, – 3) and B (5, – 8, 3).

Answer:

Let the points P(x1,y1,z1) & Q(x2,y2,z2), trisect line segment AB & let P divide the line in the ratio of 1:2 & Q in the ratio of 2:1.
Thus, P(x1,y1,z1) = P(1(5) + 2(2)/1+2, 1(-8) + 2(1)/1+2, 1(3) + 2(-3)/1+2)
= P(3,-2,-1)
& Q(x2,y2,z2) = Q(2(5) + 1(2)/2+1, 2(-8) + 1(1)/2+1, 2(3) + 1(-3)/2+1)
= Q(4,-5,1)

Question:16

If the origin is the centroid of a triangle ABC having vertices A (a, 1, 3), B (– 2, b, – 5) and C (4, 7, c), find the values of a, b, c.

Answer:

Given: Vertices of triangle ABC are-
A(z,1,3), B(-2,b,-5) & C(4,7,c)
& centroid, G = (0,0,0)
Thus, G(0,0,0) = G(a-2+4/3, 1+b+7/3, 3-5+c/3)
Thus, a+2/3 = 0
a = -2
b+8/3 = 0
b = -8
c-2/3 = 0
c = 2

Question:17

Let A (2, 2, – 3), B (5, 6, 9) and C (2, 7, 9) be the vertices of a triangle. The internal bisector of the angle A meets BC at the point D. Find the coordinates of D.

Answer:

Let us assume that the coordinates of D are (x,y,z).
Now,
AB = \sqrt{(5-2)^{2}+(6-2)^{2}+(9+3)^{2} }
= \sqrt{9+16+144 } = 13
AC =\sqrt{(2-2)^{2}+(7-2)^{2}+(9+3)^{2} }
= \sqrt{0+25+144 } = 13
Thus, AB = AC.
Thus, ABC is an isosceles triangle.
Now, D is the mid-point of BC or Angle bisector AD bisects BD,
Thus, D = (5+2/2, 6+7/2, 9+9/2)
= (7/2, 13/2,9)

Question:18

Show that the three points A (2, 3, 4), B (–1, 2, – 3) and C (– 4, 1, – 10) are collinear and find the ratio in which C divides AB.

Answer:

Given:
A(2,3,4), B(-1,2,-3) & C(-4,1,-10)
Now,
AB = \sqrt{(2+1)^{2}+(3-2)^{2}+(4+3)^{2} }
= \sqrt{9+1+49 } = \sqrt{59 }
BC = \sqrt{(-1+4)^{2}+(2-1)^{2}+(-3+10)^{2}}
= \sqrt{9+1+49 } = \sqrt{59 }
AC = \sqrt{(2+4)^{2}+(3-1)^{2}+(4+10)^{2} }
= \sqrt{36+4+196 } = 2\sqrt{59}
Now, AB + BC = \sqrt{59}+\sqrt{59}
= 2\sqrt{59}
= AC
Thus, A,B & C are collinear.AC:BC = 2\sqrt{59}:\sqrt{59} = 2:1
Thus, C divides AB externally in the ratio 2:1.

Question:19

The mid-point of the sides of a triangle are (1, 5, – 1), (0, 4, – 2) and (2, 3, 4). Find its vertices. Also find the centroid of the triangle.

Answer:

Given:
Mid-points of sides of triangle ABC are,
D(1,5,-1), E(0,4,-2) & F(2,3,4)
Let us assume that the vertices of the triangle are –
A(x1, y1, z1), B(x2, y2, z2) & C(x3, y3, z3).
Now, we know that E is the mid-point of AC, thus,
(x1 + x3/2, y1 + y3/2, z1 + z3/2) = (0,4,-2)
Thus,
C(x3, y3, z3) = C(-x1, 8 – y1, -4-z­1)
Now, mid-point of AB, F is
(x1 + x2/2, y1 + y2/2, z1 + z2/2) = (2,3,4)
Now, B(x2, y2, z2) = B(4-x1, 6-y1, 8-z1)
Now, mid-point of BC, D is,
-x1 + 4 – x1/2 = 1,
8 – y1 + 6 – y1/2 = 5
& - 4 – z1 + 8 – z1/2 = -1
Thus, x1 = 1, y1 = 2 & z1 = 3
Thus, A = (1,2,3), B = (3,4,5) & C = (-1,6,-7)
Now, for centroid,
G = (1+3-1/3,2+4+6/3, 3+5-7/3)
= (1,4,1/3)

Question:20

Prove that the points (0, – 1, – 7), (2, 1, – 9) and (6, 5, – 13) are collinear. Find the ratio in which the first point divides the join of the other two

Answer:

Given:
4(0,-1,-7), 8(2,1,-9) & C(6,5,-13)
Now,
AB = \sqrt{(0-2)^{2}+(-1-1)^{2}+(-7+9)^{2} }
\sqrt{4+4+4}=2\sqrt{3}
BC = \sqrt{(2-6)^{2}+(1-5)^{2}+(-9+13)^{2} }
\sqrt{16+16+16}=4\sqrt{3}
AC = \sqrt{(0-6)^{2}+(-1-5)^{2}+(-7+13)^{2} }
\sqrt{36+36+36}=6\sqrt{3}
Now, AB + BC = 2\sqrt{3}+4\sqrt{3}
= 6\sqrt{3}
= AC
Thus, A,B & C are collinear.
AB:AC = 2\sqrt{3}:6\sqrt{3}
= 1:3
Thus, A divides BC externally in the ratio 1:3.

Question:21

What are the coordinates of the vertices of a cube whose edge is 2 units, one of whose vertices coincides with the origin and the three edges passing through the origin, coincides with the positive direction of the axes through the origin?

Answer:

For the cube whose edge is 2 units, the coordinates will be,
(2,0,0), (2,2,0), (0,2,0), (0,2,2), (0,0,2), (2,0,2), (0,0,0) & (2,2,2).

Question:22

The distance of point P(3, 4, 5) from the yz-plane is
A. 3 units
B. 4 units
C. 5 units
D. 550

Answer:

The answer is the option (a) 3 units
Given: P(3,4,5)
Now, from yz-plane, distance of P,
=| x coordinate of P| = 3

Question:23

What is the length of foot of perpendicular drawn from the point P (3, 4, 5) on y-axis
A. \sqrt{41}
B. \sqrt{34}
C. 5
D. None of these

Answer:

The answer is the option (b)\sqrt{ 34}
On y-axis, x = z = 0
Thus, A = (0,4,0)
Thus, PA = \sqrt{ (0-3)^{2}+(4-4)^{2}+(0-5)^{2} }
= \sqrt{ 9+0+25 } = \sqrt{ 34 }

Question:24

Distance of the point (3, 4, 5) from the origin (0, 0, 0) is
A. \sqrt{50}
B. 3
C. 4
D. 5

Answer:

The answer is the option (a) \sqrt{50}
Given: P(3,4,5) & O(0,0,0)
Thus, OP = \sqrt{(0-3)^{2}+(0-4)2^{+}(0-5)^{2} }
= \sqrt{9+16+25} =\sqrt{50}

Question:25

If the distance between the points (a, 0, 1) and (0, 1, 2) is \sqrt{27} , then the value of a is
A. 5
B. ± 5
C. – 5
D. none of these

Answer:

The answer is the option (b) ±5
Given: A(a,0,1) & B(0,1,2)
& OP = \sqrt{(a-0)^{2}+(0-1)^{2}+(1-2)^{2}}
= \sqrt{27}
Thus, a2 + 2 = 27
a2 = 25
a = ±5

Question:26

x-axis is the intersection of two planes
A. xy and xz
B. yz and zx
C. xy and yz
D. none of these

Answer:

The answer is the option (a) xy & xz
ON xy-plane & xz-plane, x-axis is the line of intersection

Question:27

Equation of y-axis is considered as
A. x = 0, y = 0
B. y = 0, z = 0
C. z = 0, x = 0
D. none of these

Answer:

The answer is the option (c) z = 0, x = 0
We know that, on y-axis,
x = 0 & z = 0.

Question:28

The point (–2, –3, –4) lies in the
A. First octant
B. Seventh octant
C. Second octant
D. Eighth octant

Answer:

The answer is the option (b) seventh octant
The given point (-2,-3,-4) lies in the seventh octant

Question:29

A plane is parallel to yz-plane so it is perpendicular to :
A. x-axis
B. y-axis
C. z-axis
D. none of these

Answer:

The answer is the option (a) x-axis
If a plane is parallel to yz-plane, then it is perpendicular to the x-axis.

Question:30

The locus of a point for which y = 0, z = 0 is
A. equation of x-axis
B. equation of y-axis
C. equation at z-axis
D. none of these
Answer:

The answer is the option (a) equation of x-axis
The locus of point is the equation of x-axis since we know that on x-axis,
y = 0 & z = 0

Question:31

The locus of a point for which x = 0 is
A. xy-plane
B. yz-plane
C. zx-plane
D. none of these

Answer:

The answer is the option (b) yz-plane
We know that, on yz-plane,
x = 0
Thus, the locus of point is yz-plane

Question:32

If a parallelepiped is formed by planes drawn through the points (5, 8, 10) and (3, 6, 8) parallel to the coordinate planes, then the length of diagonal of the parallelepiped is
A. 2\sqrt{3}
B. 3\sqrt{2}
C. \sqrt{2}
D. \sqrt{3}

Answer:

The answer is the option (a) 2\sqrt{3}
Given: The parallelepiped passes through A(5,8,10) & B(3,6,8)
Thus,
Length of the diagonal, AB = \sqrt{(5-3)^{2}+(8-6)^{2}+(10-8)^{2} }
= \sqrt{4+4+4}=2\sqrt{3}

Question:33

L is the foot of the perpendicular drawn from a point P (3, 4, 5) on the xy-plane. The coordinates of point L are
A. (3, 0, 0)
B. (0, 4, 5)
C. (3, 0, 5)
D. none of these

Answer:

The answer is the option (d) none of these
On xy-plane, z = 0.
Thus, coordinates of L are (3,4,0).

Question:34

L is the foot of the perpendicular drawn from a point (3, 4, 5) on x-axis. The coordinates of L are
A. (3, 0, 0)
B. (0, 4, 0)
C. (0, 0, 5)
D. none of these

Answer:

The answer is the option (a) (3,0,0)
We know that, on x-axis,
y = z = 0.
Thus, the coordinates will be, (3,0,0)

Question:35

Fill in the blanks
The three axes OX, OY, OZ determine ________ .

Answer:

coordinate planes.

Question:36

Fill in the blanks
The three planes determine a rectangular parallelopiped which has ________ of rectangular faces.

Answer:


Here, the rectangular parallelepiped is determined by the three planes, ABB’A, AA’D’D, A’B’C’D’.
Here, we have 3 pairs of rectangular faces –
(ABB’A, DCC’D’), (ABCD, A’B’C’D’), (ADD’A’, BCC’B’)

Question:37

Fill in the blanks
The coordinates of a point are the perpendicular distance from the ________on the respective axes.

Answer:

Given points

Question:38

Fill in the blanks
The three coordinate planes divide the space into ________ parts

Answer:

Eight

Question:39

Fill in the blanks
If a point P lies in yz-plane, then the coordinates of a point on yz-plane is of the form ________.

Answer:

On yz-plane, x = 0.
Thus, the coordinates will be – (0,y,z).

Question:40

Fill in the blanks
The equation of yz-plane is ________.

Answer:

On yz-plane, x = 0.
Thus, the equation of yz-plane is x = 0

Question:41

Fill in the blanks
If the point P lies on z-axis, then coordinates of P are of the form ________.

Answer:

We know that,
On z-axis, x = y = 0.
Thus, the coordinates are (0,0,z).

Question:42

Fill in the blanks
The equation of z-axis, are ________.

Answer:

On z-axis any point will be (0,0,z).
Thus, On z-axis, x = y = 0, viz. represented in the equation.

Question:43

Fill in the blanks
A line is parallel to xy-plane if all the points on the line have equal ________.

Answer:

If each point P(x,y,z) is at the same distance from xy-plane, then the line is parallel to xy-plane.
Thus, distance of xy-plane from point P = z
Thus, if all the points on the line have equal z coordinates then the line is parallel to xy-plane.

Question:44

Fill in the blanks
A line is parallel to x-axis if all the points on the line have equal ________.

Answer:

If each point on a line maintains a constant distance from y-axis and z-axis, then it is parallel to x-axis.
Therefore, each point has equal y & z co-ordinates.

Question:45

Fill in the blanks
x = a represent a plane parallel to ________.

Answer:

x = a is the locus of point P(x,y,z).
Thus, each point P has a constant x-coordinate.
Also, distance of yz-plane from P is ‘x’.
Thus, plane x = a is parallel to yz-plane & is at a constant distance ‘a’ from yz-plane.

Question:46

Fill in the blanks
The plane parallel to yz - plane is perpendicular to ________.

Answer:

x-axis

Question:47

Fill in the blanks
The length of the longest piece of a string that can be stretched straight in a rectangular room whose dimensions are 10, 13 and 8 units are ______.

Answer:

Given:
a = 10, b = 13 & c = 8.
Now, length of the string –
\\ =\sqrt{a^2+b^2+c^2} \\= \sqrt{10^2+13^2+8^2}\\ =\sqrt{100+169+64}\\ = \sqrt{333}

Question:48

Fill in the blanks
If the distance between the points (a, 2, 1) and (1, –1, 1) is 5, then a _______.

Answer:

Given points: (a,2,1,) & (1,-1,1)
\sqrt{(a-1)^{2}+(2+1)^{2}+(1-1)^{2}} =5
(a-1)2 + 9 + 0 = 25
a2 – 2a – 15 = 0
(a-5)(a+3) = 0
Thus, a = -3 or a = 5

Question:49

Fill in the blanks
If the mid-points of the sides of a triangle AB; BC; CA are D (1, 2, – 3), E (3, 0, 1) and F (–1, 1, – 4), then the centroid of the triangle ABC is ________.

Answer:

Given:
Mid-points of triangle ABC are D(1,2,-3), E(3,0,1) & F(-1,1,-4).
Now, the centroid of triangle DEF = Centroid of triangle ABC ……. (from geometry of centroid)
Thus, centroid of triangle ABC,
G(1+3-1/3, 2+0+1/3, -3+1-4/3) ≡ G(1,1,-2)

Question:50

Match each item given under the column C1 to its correct answer given under column C2.

Column C,

Column C2

(a)

In xy-plane

(i)

1st octant

(b)

Point (2, 3,4) lies in the

(ii)

vz-plane

(c)

Locus of the points having x coordinate 0 is

(iii)

z-coordinate is zero

(d)

A line is parallel to x-axis if and only

(iv)

z-axis .

(e)

If x = 0, y = 0 taken together will represent the

(v)

plane parallel to xy-plane

(f)

z = c represent the plane

(vi)

if all the points on the line have equal y and z-coordinates.

(g)

Planes x = a, y = b represent the line

(vii)

from the point on the respective axis.

00

Coordinates of a point are the distances from the origin to the feet of perpendiculars

(viii)

parallel to z-axis

(i)

A ball is the solid region in the space

(ix)

disc

G)

Region in the plane enclosed by a circle is known as a

(x)

sphere

Answer:

Solution:
(a) → (iii)
In xy-plane, z = 0.
(b)→(i)
Point (2,3,4) lies in the first octant.
(c) → (ii)
Yz-plane is the locus of points having, x = 0
(d) → (vi)
Only if all the points on a line have equal y & z co-ordinates, the line will be parallel to x-axis.
(e) → (iv)
z-axis is represented by x = y = 0.
(f) → (v)
Plane parallel to xy-plane is represented by z = c.
(g) → (viii)
Planes x = a & y = b is a line of intersection of these planes, since x = a is parallel to yz-plane & y = b is parallel to xz-plane.
z-axis is the line of intersection of yz & xz planes.
Thus, the line of intersection is parallel to z-axis.
(h) → (vii)
The coordinate of a point can be defined as the distance from the origin to the feet of perpendicular from the point on their respective axis.

  1. → (x)

A ball can be said the solid region in the space enclosed by the sphere.
(j) → (ix)
A disc is the region in a plane enclosed by a circle.

More about NCERT Exemplar Class 11 Maths Solutions Chapter 12

The solutions for NCERT problems can be easily accessed offline through NCERT Exemplar Class 11 Maths solutions chapter 12 PDF download for a better understanding of the concepts of Three Dimensional geometry that are formed efficiently through the help of professionals.

The NCERT Exemplar solutions for Class 11 Maths chapter 12 provides students with the best possible and carefully examined solutions that make the learning process interesting and knowledgeable at the same time.

Topics and Subtopics in NCERT Exemplar Class 11 Maths Solutions Chapter 12

Introduction to Three Dimensional Geometry is the 12th chapter of maths which contain the following topics:

  • 12.1 Introduction
  • 12.2 Coordinate axis and coordinate planes in three-dimension space
  • 12.3 Coordinates of a point in space
  • 12.4 Distance between two points
  • 12.5 Section formula

What will the Students Get To Know in NCERT Exemplar Class 11 Maths Solutions Chapter 12?

The learners will be introduced to a new system of three-dimensional geometry along with a detailed explanation of its basics, including the study of coordinate axes and planes in three-dimensional space. The learners will also get a detailed representation of three-dimensional coordinate planes into eight octants along with their sign convention. The students will also learn to calculate the coordinates and distance of any two points in three-dimensional space.

NCERT Exemplar Class 11 Maths Solutions Chapter-Wise

Important Topics To Cover From NCERT Exemplar Solutions For Class 11 Maths Chapter 12

  • The students will be introduced to three-dimensional geometry along with its basic concepts and properties.
  • The students will learn about coordinate axes and planes of three-dimensional space along with their representation.
  • NCERT Exemplar Class 11 Maths solutions chapter 12 also covers the calculation and determination of coordinates of a point in space.
  • The students should cover the usage of the distance formula for the calculation of the distance between any two points given in the three-dimensional space.
  • NCERT Exemplar Class 11 Maths chapter 12 solutions also provides the derivation and formula for finding the coordinates of any given point through the use of section formula that is important to study from the perspective of academic exams as well as competitive exams.
  • The students should practice all the NCERT solutions and examples repeatedly for the exams.

Check Chapter-Wise NCERT Solutions of Book

NCERT Exemplar Class 11 Solutions

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Frequently Asked Question (FAQs)

1. Is the chapter important from the perspective of Board exams?

Yes, the chapter is very important for Board exams along with competitive exams as this chapter has high weightage in both examinations.

2. Can I download the solutions of this chapter for offline access?

Yes, the students can download the NCERT Exemplar Class 11 Maths Solutions Chapter 12 provided on our website through online tools available for downloading web pages as pdf.

3. What are the topics covered in this chapter?

There are the topics covered in this chapter

  • 12.1 Introduction
  • 12.2 Coordinate axis and coordinate planes in three-dimension space
  • 12.3 Coordinates of a point in space
  • 12.4 Distance between two points
  • 12.5 Section formula

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Careers in computer programming primarily refer to the systematic act of writing code and moreover include wider computer science areas. The word 'programmer' or 'coder' has entered into practice with the growing number of newly self-taught tech enthusiasts. Computer programming careers involve the use of designs created by software developers and engineers and transforming them into commands that can be implemented by computers. These commands result in regular usage of social media sites, word-processing applications and browsers.

3 Jobs Available
Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.  

3 Jobs Available
Information Security Manager

Individuals in the information security manager career path involves in overseeing and controlling all aspects of computer security. The IT security manager job description includes planning and carrying out security measures to protect the business data and information from corruption, theft, unauthorised access, and deliberate attack 

3 Jobs Available
ITSM Manager
3 Jobs Available
Automation Test Engineer

An Automation Test Engineer job involves executing automated test scripts. He or she identifies the project’s problems and troubleshoots them. The role involves documenting the defect using management tools. He or she works with the application team in order to resolve any issues arising during the testing process. 

2 Jobs Available

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