Cartesian Product: Definition, Illustrations, and Solved Examples for Class 10 Maths`

Cartesian Product - Definition, Illustration, Example, Solution

Cartesian Product

Illustration 1

Let us consider the following two sets.

A is the set of 3 vegetables and B is the set of 4 fruits. That is,

A = {carrot, brinjal, ladies finger} and B = {apple, orange, grapes, strawberry}

What are the possible ways of choosing a vegetable with a fruit? (Fig.1.2)

Fig. 1.2

We can select them in 12 distinct pairs as given below.

(c, a), (c, o), (c, g), (c, s), (b, a), (b, o), (b, g), (b, s), (l, a), (l, o), (l,g), (l, s)

This collection represents the cartesian product of the set of vegetables and set of fruits.

Definition

If A and B are two non-empty sets, then the set of all ordered pairs (a, b) such that \( a \in A \), \( b \in B \) is called the Cartesian Product of A and B, and is denoted by \( A \times B \). Thus, \( A \times B = \{(a,b) \mid a \in A, b \in B\} \).

Note

• \( A \times B \) is the set of all possible ordered pairs between the elements of A and B such that the first coordinate is an element of A and the second coordinate is an element of B.
• \( B \times A \) is the set of all possible ordered pairs between the elements of A and B such that the first coordinate is an element of B and the second coordinate is an element of A.
• If \( a = b \), then \( (a, b) = (b, a) \).
• The “cartesian product” is also referred as “cross product”.

Illustration 2

Let A = {1, 2, 3} and B = {a, b}. Write \( A \times B \) and \( B \times A \) ?

\( A \times B = \{1,2,3\} \times \{a,b\} = \{(1, a ),(1, b ),(2, a ),(2, b ),(3, a ),(3, b )\} \) (as shown in Fig.1.3)

\( B \times A = \{a,b\} \times \{1,2,3\} = \{(a,1), (a,2), (a,3),(b,1), (b,2), (b,3)\} \) (as shown in Fig.1.3)

Fig. 1.3

Recall of standard infinite sets

Natural Numbers N = {1, 2, 3, 4…}

Whole Numbers W = {0,1,2,3, ...}

Integers Z ={..., –2,–1,0,1,2, ...}

Rational Numbers \( \mathbf{Q} = \{ \frac{p}{q} \mid p, q \in \mathbf{Z}, q \neq 0 \} \)

Real Numbers \( \mathbf{R} = \mathbf{Q} \cup \mathbf{Q}’ \), where \( \mathbf{Q}’ \) is the set of all irrational numbers.

Illustration 3

For example, let A be the set of numbers in the interval [3, 5] and B be the set of numbers in the interval [2,3]. Then the Cartesian product \( A \times B \) corresponds to the rectangular region shown in the Fig. 1.4. It consists of all points (x, y) within the region.

Fig. 1.4

Progress check

1. For any two non-empty sets A and B, \( A \times B \) is called as ______.
2. If \( n(A \times B) = 20 \) and \( n(A) = 5 \) then \( n(B) \) is ______.
3. If \( A = \{-1,1\} \) and \( B = \{-1,1\} \) then geometrically describe the set of points of \( A \times B \).
4. If A, B are the line segments given by the intervals (–4, 3) and (–2, 3) respectively, represent the cartesian product of A and B.

Note:

The set of all points in the cartesian plane can be viewed as the set of all ordered pairs (x, y) where x, y are real numbers. In fact, \( \mathbb{R} \times \mathbb{R} \) is the set of all points which we call as the cartesian plane.

Example 1.1

If \( A = \{1,3,5\} \) and \( B = \{2,3\} \) then (i) find \( A \times B \) and \( B \times A \).

(ii) Is \( A \times B = B \times A \)? If not why?

(iii) Show that \( n(A \times B) = n(B \times A) = n(A) \times n(B) \).

Solution

Given that \( A = \{1,3,5\} \) and \( B = \{2,3\} \)

(i) \( A \times B = \{1,3,5\} \times \{2,3\} = \{(1,2), (1,3), (3,2), (3,3), (5,2), (5,3)\} \) ...(1)

\( B \times A = \{2,3\} \times \{1,3,5\} = \{(2,1), (2,3), (2,5), (3,1), (3,3), (3,5)\} \) ...(2)

(ii) From (1) and (2) we conclude that \( A \times B \neq B \times A \) as \( (1, 2) \neq (2, 1) \) and \( (1, 3) \neq (3, 1) \), etc.

(iii) \( n(A)=3 \); \( n(B) = 2 \).

From (1) and (2) we observe that, \( n(A \times B) = n(B \times A) = 6 \);

we see that, \( n(A) \times n(B) = 3 \times 2 = 6 \) and \( n(B) \times n(A) = 2 \times 3 = 6 \)

Hence, \( n(A \times B) = n(B \times A) = n(A) \times n(B) = 6 \).

Thus, \( n(A \times B) = n(B \times A) = n(A) \times n(B) \).

Example 1.2

If \( A \times B = \{(3,2), (3,4), (5,2), (5,4)\} \) then find A and B.

Solution

\( A \times B = \{(3,2), (3,4), (5,2), (5,4)\} \)

We have A = {set of all first coordinates of elements of \( A \times B \)}. Therefore, A = {3,5}

B = {set of all second coordinates of elements of \( A \times B \)}. Therefore, B = {2,4}

Thus \( A = \{3,5\} \) and \( B = \{2,4\} \).

Example 1.3

Let \( A = \{x \in \mathbf{N} \mid 1 < x < 4\} \), \( B = \{x \in \mathbf{W} \mid 0 \leq x < 2\} \) and \( C = \{x \in \mathbf{N} \mid x < 3\} \).

Then verify that

(i) \( A \times (B \cup C) = (A \times B) \cup (A \times C) \)

(ii) \( A \times (B \cap C) = (A \times B) \cap (A \times C) \)

Solution

\( A = \{x \in \mathbf{N} \mid 1 < x < 4\} = \{2, 3\} \),

\( B = \{x \in \mathbf{W} \mid 0 \leq x < 2\} = \{0, 1\} \),

\( C = \{x \in \mathbf{N} \mid x < 3\} = \{1, 2\} \)

(i) \( A \times (B \cup C) = (A \times B) \cup (A \times C) \)

\( B \cup C = \{0, 1\} \cup \{1, 2\} = \{0, 1, 2\} \)

\( A \times (B \cup C) = \{2, 3\} \times \{0, 1, 2\} = \{(2, 0), (2, 1), (2, 2), (3, 0), (3, 1), (3, 2)\} \) ...(1)

\( A \times B = \{2, 3\} \times \{0, 1\} = \{(2,0),(2,1),(3,0),(3,1)\} \)

\( A \times C = \{2, 3\} \times \{1, 2\} = \{(2, 1), (2, 2), (3, 1), (3, 2)\} \)

\( (A \times B) \cup (A \times C) = \{(2, 0), (2, 1), (3, 0), (3, 1)\} \cup \{(2, 1), (2, 2), (3, 1), (3, 2)\} \)

\( = \{(2, 0), (2, 1), (2, 2), (3, 0), (3, 1), (3, 2)\} \) ...(2)

From (1) and (2), \( A \times (B \cup C) = (A \times B) \cup (A \times C) \) is verified.

(ii) \( A \times (B \cap C) = (A \times B) \cap (A \times C) \)

\( B \cap C = \{0, 1\} \cap \{1, 2\} = \{1\} \)

\( A \times (B \cap C) = \{2, 3\} \times \{1\} = \{(2,1),(3,1)\} \) ... (3)

\( A \times B = \{2, 3\} \times \{0, 1\} = \{(2, 0),(2, 1),(3, 0),(3, 1)\} \)

\( A \times C = \{2, 3\} \times \{1, 2\} = \{(2, 1),(2, 2),(3, 1),(3, 2)\} \)

\( (A \times B) \cap (A \times C) = \{(2, 0),(2, 1),(3, 0),(3, 1)\} \cap \{(2, 1),(2, 2),(3, 1),(3, 2)\} \)

\( = \{(2, 1),(3, 1)\} \) ... (4)

From (3) and (4), \( A \times (B \cap C) = (A \times B) \cap (A \times C) \) is verified.

Note

The above two verified properties are called distributive property of cartesian product over union and intersection respectively. In fact, for any three sets A, B, C we have

(i) \( A \times (B \cup C) = (A \times B) \cup (A \times C) \)

(ii) \( A \times (B \cap C) = (A \times B) \cap (A \times C) \)

Cartesian Product of three Sets

If A, B, C are three non-empty sets then the cartesian product of three sets is the set of all possible ordered triplets given by

\( A \times B \times C = \{(a,b,c) \text{ for all } a \in A, b \in B, c \in C \} \)

Illustration for Geometrical understanding of cartesian product of two and three sets

Let \( A = \{0,1\} \), \( B = \{0,1\} \), \( C = \{0,1\} \)

\( A \times B = \{0,1\} \times \{0,1\} = \{(0, 0),(0,1),(1, 0),(1,1)\} \)

Fig. 1.5

Representing \( A \times B \) in the xy - plane we get a picture shown in Fig. 1.5.

\( (A \times B) \times C = \{(0, 0),(0,1),(1, 0),(1,1)\} \times \{0,1\} \)

\( = \{(0, 0, 0),(0, 0,1),(0,1, 0),(0,1,1),(1, 0, 0),(1, 0,1)(1,1, 0),(1,1,1)\} \)

Representing \( A \times B \times C \) in the xyz - plane we get a picture as shown in Fig. 1.6

Fig. 1.6

Thus, \( A \times B \) represent vertices of a square in two dimensions and \( A \times B \times C \) represent vertices of a cube in three dimensions.

NOTES

In general, cartesian product of two non-empty sets provides a shape in two dimensions and cartesian product of three non-empty sets provide an object in three dimensions.

Solved Problems: Atoms and Molecules | Chemistry & Science Guide

Solved Problems: Atoms and Molecules

Table of Contents

1. Calculation of molar mass
2. Calculation based on number of moles from mass and volume
3. Calculation of mass from mole
4. Calculation based on number of atoms/molecules
5. Calculation based on molar volume
6. Calculation based on % composition

I. Calculation of molar mass

Calculate the gram molar mass of the following.

1) H₂O        2) CO₂        3) Ca₃(PO₄)₂

Solution:

1) H₂O

Atomic masses of H = 1, O = 16

Gram molar mass of H₂O = (1 × 2) + (16 × 1)

= 2 + 16

Gram molar mass of H₂O = 18 g

2) CO₂

Atomic masses of C = 12, O = 16

Gram molar mass of CO₂ = (12 × 1) + (16 × 2)

= 12 + 32

Gram molar mass of CO₂ = 44 g

3) Ca₃(PO₄)₂

Atomic masses of Ca = 40, P = 30, O = 16.

Gram molar mass of Ca₃(PO₄)₂ = (40 × 3) + [30 + (16 × 4)] × 2

= 120 + (94 × 2)

= 120 + 188

Gram molar mass of Ca₃(PO₄)₂ = 308 g

II. Calculation based on number of moles from mass and volume

1) Calculate the number of moles in 46 g of sodium?

= 46 / 23

2 moles of sodium

2) 5.6 litre of oxygen at S.T.P

3) Calculate the number of moles of a sample that contains 12.046 × 10²³ atoms of iron ?

III. Calculation of mass from mole

Calculate the mass of the following

1) 0.3 mole of aluminium (Atomic mass of Al = 27)

Mass = No. of moles × atomic mass

So, mass of Al = 0.3 × 27

= 8.1 g

2) 2.24 litre of SO₂ gas at S.T.P

Molecular mass of SO₂ = 32 + (16 × 2) = 32 + 32 = 64

Mass = No. of moles × molecular mass

Mass = 0.1 × 64

Mass of SO₂ = 6.4 g

3) 1.51 × 10²³ molecules of water

Molecular mass of H₂O = 18

0.25 = mass / 18

Mass = 0.25 × 18

Mass = 4.5 g

4) 5 × 10²³ molecules of glucose ?

Molecular mass of glucose = 180

= (180 × 5 × 10²³) / 6.023 × 10²³

= 149.43 g

IV. Calculation based on number of atoms/ molecules.

1) Calculate the number of molecules in 11.2 litre of CO₂ at S.T.P

= 11.2 / 22.4

= 0.5 mole

Number of molecules of CO₂ = number of moles of CO₂ × Avogadro’s number

= 0.5 × 6.023 × 10²³

= 3.011 × 10²³ molecules of CO₂

2) Calculate the number of atoms present in 1 gram of gold (Atomic mass of Au = 198)

Number of atoms of Au = 3.042 × 10²¹ g

3) Calculate the number of molecules in 54 gm of H₂O?

Number of molecules of water = 6.023 × 10²³ × 54 / 18

= 18.069 × 10²³ molecules

4) Calculate the number of atoms of oxygen and carbon in 5 moles of CO₂.

1 mole of CO₂ contains 2 moles of oxygen

5 moles of CO₂ contain 10 moles of oxygen

Number of atoms of oxygen = Number of moles of oxygen × Avogadro’s number

= 10 × 6.023 × 10²³

= 6.023 × 10²⁴ atoms of Oxygen

1 mole of CO₂ contains 1 mole of carbon

5 moles of CO₂ contains 5 moles of carbon

No. of atoms of carbon = No.of moles of carbon × Avogadro’s number

= 5 × 6.023 × 10²³

= 3.011 × 10²⁴ atoms of Carbon

V. Calculation based on molar volume

Calculate the volume occupied by:

1) 2.5 mole of CO₂ at S.T.P

2) 3.011 × 10²³ of ammonia gas molecules

= 2 moles

Volume occupied by NH₃ = number of moles × molar volume

= 2 × 22.4

= 44.8 litres at S.T.P

3) 14 g nitrogen gas

Number of moles = 14 / 28

= 0.5 mole

Volume occupied by N₂ at S.T.P = no. of moles × molar volume

0.5 × 22.4

11.2 litres.

VI. Calculation based on % composition

Calculate % of S in H₂SO₄

Molar mass of H₂SO₄ = (1 × 2) + (32 × 1) + (16 × 4)

= 2 + 32 + 64

= 98 g

Comprehensive Solved Problems in Nuclear Physics for Class 10 Science

Solved problem - Nuclear Physics | Science

Solved problem 6.1

Identify A, B, C, and D from the following nuclear reactions.

Solution:

A is alpha particle, B is neutron, C is proton, and D is electron.

Solved problem 6.2

A radon specimen emits radiation of \(3.7 \times 10^3\) GBq per second. Convert this disintegration in terms of curie. (one curie = \(3.7 \times 10^{10}\) disintegration per second)

Solution:

1 Bq = one disintegration per second one curie = \(3.7 \times 10^{10}\) Bq

$$ \text{Given radiation} = 3.7 \times 10^3 \text{ GBq} $$ $$ = 3.7 \times 10^3 \times 10^9 \text{ Bq} $$ $$ = 3.7 \times 10^{12} \text{ Bq} $$ $$ \text{Since 1 curie} = 3.7 \times 10^{10} \text{ Bq} $$ $$ \text{Therefore, radiation in curie} = \frac{3.7 \times 10^{12}}{3.7 \times 10^{10}} $$ $$ = 100 \text{ curie} $$

Solved problem 6.3

\(^{235}_{92}U\) experiences one \(\alpha -\)decay and one \(\beta -\) decay. Find number of neutrons in the final daughter nucleus that is formed.

Solution:

Let X and Y be the resulting nucleus after the emission of the alpha and beta particles respectively.

$$ ^{235}_{92}U \xrightarrow{\alpha \text{-decay}} ^{231}_{90}X + ^{4}_{2}He $$ $$ ^{231}_{90}X \xrightarrow{\beta \text{-decay}} ^{231}_{91}Y + e^{-} $$

Number of neutrons = Mass number – Atomic number = 231 – 91 = 140

Solved problem 6.4

Calculate the amount of energy released when a radioactive substance undergoes fusion and results in a mass defect of 2 kg.

Solution:

Mass defect in the reaction (m) = 2 kg

Velocity of light (c) = \(3 \times 10^8 \text{ m s}^{-1}\)

By Einstein’s equation,

$$ E = mc^2 $$

So E = \(2 \times (3 \times 10^8)^2\)

= \(1.8 \times 10^{17}\) J

35 Solved Coordinate Geometry Questions for Class 10 Math

35 Solved Coordinate Geometry Questions

1. Find the distance between (2, 3) and (4, 1).

$d = \sqrt{(4-2)^2 + (1-3)^2} = \sqrt{2^2 + (-2)^2} = \sqrt{8} = $ 2√2 units

2. Find the distance between (-5, 7) and (-1, 3).

$d = \sqrt{(-1 - (-5))^2 + (3-7)^2} = \sqrt{4^2 + (-4)^2} = \sqrt{32} = $ 4√2 units

3. Distance of point (x, y) from origin (0, 0)?

Distance = $\sqrt{x^2 + y^2}$

4. Find distance between (a, b) and (-a, -b).

$d = \sqrt{(-a-a)^2 + (-b-b)^2} = \sqrt{(-2a)^2 + (-2b)^2} = \sqrt{4a^2 + 4b^2} = $ 2√(a² + b²)

5. Are points (1, 5), (2, 3) and (-2, -11) collinear?

Check if Area of Triangle = 0. $1(3 - (-11)) + 2(-11 - 5) + (-2)(5 - 3) = 14 - 32 - 4 = -22 \neq 0$.
Result: Not Collinear.

6. Find midpoint of (6, 8) and (2, 4).

$M = (\frac{6+2}{2}, \frac{8+4}{2}) = (4, 6)$.

7. Find 'y' if distance between (2, -3) and (10, y) is 10.

$10^2 = (10-2)^2 + (y+3)^2 \Rightarrow 100 = 64 + (y+3)^2 \Rightarrow 36 = (y+3)^2$.
$y+3 = \pm 6 \Rightarrow y = 3$ or $y = -9$.

8. Relation between x and y such that (x, y) is equidistant from (7, 1) and (3, 5)?

$(x-7)^2 + (y-1)^2 = (x-3)^2 + (y-5)^2$. Simplifying: x - y = 2.

9. Point on x-axis equidistant from (2, -5) and (-2, 9).

Point is (x, 0). $(x-2)^2 + (0+5)^2 = (x+2)^2 + (0-9)^2 \Rightarrow -8x = 56 \Rightarrow x = -7$.
Point: (-7, 0).

10. Find the centroid of triangle with vertices (1,1), (0,0), and (2,2).

$G = (\frac{1+0+2}{3}, \frac{1+0+2}{3}) = (1, 1)$.

11. Find coordinates dividing (-1, 7) and (4, -3) in ratio 2:3.

$x = \frac{2(4)+3(-1)}{5} = 1, y = \frac{2(-3)+3(7)}{5} = 3$. Point: (1, 3).

12. Coordinates of points of trisection of (4, -1) and (-2, -3).

Ratio 1:2 $\Rightarrow$ (2, -5/3); Ratio 2:1 $\Rightarrow$ (0, -7/3).

13. Find ratio in which y-axis divides (-4, 5) and (3, -7).

On y-axis, x=0. $0 = \frac{k(3) + 1(-4)}{k+1} \Rightarrow 3k = 4 \Rightarrow k = 4:3$.

14. Ratio in which (-3, 10) and (6, -8) is divided by (-1, 6).

$-1 = \frac{6k - 3}{k+1} \Rightarrow -k-1 = 6k-3 \Rightarrow 7k=2 \Rightarrow$ 2:7.

15. Area of triangle with vertices (2, 3), (-1, 0), (2, -4).

$\frac{1}{2}|2(0 - (-4)) + (-1)(-4 - 3) + 2(3 - 0)| = \frac{1}{2}|8 + 7 + 6| = $ 10.5 sq. units.

16. Find 'k' if (2, 3), (4, k), (6, -3) are collinear.

Area = 0 $\Rightarrow 2(k+3) + 4(-3-3) + 6(3-k) = 0 \Rightarrow -4k = 0 \Rightarrow k = 0$.

17. Perimeter of triangle with vertices (0,4), (0,0), (3,0).

Sides: 4, 3, and $\sqrt{3^2+4^2}=5$. Perimeter = $4+3+5 = 12$ units.

18. Find fourth vertex of parallelogram: (1,2), (4,3), (6,6).

Midpoints of diagonals are same. $(1+6)/2 = (4+x)/2 \Rightarrow x=3$. $(2+6)/2 = (3+y)/2 \Rightarrow y=5$. Vertex: (3, 5).

19. Coordinate of point A where AB is diameter, center is (2, -3), B is (1, 4).

$2 = (x+1)/2 \Rightarrow x=3$. $-3 = (y+4)/2 \Rightarrow y=-10$. A is (3, -10).

20. Distance between $(a\cos\theta, 0)$ and $(0, a\sin\theta)$.

$\sqrt{(0-a\cos\theta)^2 + (a\sin\theta-0)^2} = \sqrt{a^2(\cos^2\theta + \sin^2\theta)} = \sqrt{a^2} = a$.

21. If P(x,y) is equidistant from A(5,1) and B(-1,5), find relation.

$(x-5)^2 + (y-1)^2 = (x+1)^2 + (y-5)^2 \Rightarrow 3x = 2y$.

22. Find area of rhombus if vertices are (3,0), (4,5), (-1,4), (-2,-1).

$\frac{1}{2} \times d_1 \times d_2$. $d_1 = \sqrt{32}, d_2 = \sqrt{72}$. Area = 24 sq. units.

23. Point P divides AB in ratio 1:3. A(2,1), B(7,6). Find P.

$x = \frac{1(7)+3(2)}{4} = 3.25, y = \frac{1(6)+3(1)}{4} = 2.25$.

24. Find distance of (4, -3) from x-axis.

Distance from x-axis = $|y| = |-3| = 3$ units.

25. Find distance of (4, -3) from y-axis.

Distance from y-axis = $|x| = 4$ units.

26. If points (0,0), (3,√3) and (3,p) form equilateral triangle, find p.

Distance OP = PQ. $p = -\sqrt{3}$.

27. Midpoint of line joining (2a, 4) and (-2, 3b) is (1, 2a+1). Find a, b.

$(2a-2)/2 = 1 \Rightarrow a=2$. $(4+3b)/2 = 2(2)+1 \Rightarrow b=2$.

28. Find area of triangle with vertices (a, b+c), (b, c+a), (c, a+b).

$Area = 0$ (Points are collinear).

29. Find distance between $(L, M)$ and $(L+a, M+b)$.

$d = \sqrt{(L+a-L)^2 + (M+b-M)^2} = \sqrt{a^2+b^2}$.

30. Coordinate of point on y-axis equidistant from (6,5) and (-4,3).

Point (0, y). $6^2 + (y-5)^2 = (-4)^2 + (y-3)^2 \Rightarrow y = 9$. Point: (0, 9).

31. Ratio in which line segment joining (1,-5) and (-4,5) is divided by x-axis.

$y=0 \Rightarrow 0 = \frac{k(5) + 1(-5)}{k+1} \Rightarrow k=1$. Ratio: 1:1.

32. Centroid of triangle with vertices (a,b), (b,c), (c,a) is origin. Find a+b+c.

$(a+b+c)/3 = 0 \Rightarrow a+b+c = 0$.

33. Distance between $(\cos\theta, \sin\theta)$ and $(\sin\theta, \cos\theta)$.

$\sqrt{(\sin\theta-\cos\theta)^2 + (\cos\theta-\sin\theta)^2} = \sqrt{2(\sin\theta-\cos\theta)^2} = \sqrt{2}|\sin\theta-\cos\theta|$.

34. Find 'p' if (p, 2) is midpoint of (5, 3) and (1, 1).

$p = (5+1)/2 = 3$.

35. Type of triangle formed by (1,1), (-1,-1), (√3, -√3).

$AB = \sqrt{8}, BC = \sqrt{8}, AC = \sqrt{8}$. Result: Equilateral Triangle.