Vector Geometry

Vector Geometry

Vector Geometry: We can study different properties and relations relating to geometry with the help of vectors. Such a study is known as the Vector Geometry.

Here are 12 geometrical theorems and their proofs by vector method.

Product of Vectors

Product of Vectors

Product of Vectors: Product of vectors or multiplication of vectors can be performed by the following two ways:

  • Scalar Product or Dot Product
  • Vector Product or Cross Product

Vector Operations

Vector Operations

Vector Operations: The vectors can be added, subtracted or multiplied with one another. Addition of vectors, difference of vectors, multiplication of a vector by a scalar, product of vectors, etc. are called vector operations.

 

Types of Vectors

Types of Vectors

Types of Vectors: After we have learnt ‘Vectors and its Components’ and ‘Magnitude and Direction of a Vector’, we’ll learn the Types of Vectors here. Vectors are of different types according to their magnitude, direction, position with respect to other given vector, and by the method of representing vectors.

Following are the different types of vectors:


Magnitude and Direction of a Vector

Magnitude and Direction of a Vector

Magnitude of a Vector 

Magnitude of a vector is the length of its directed line segment from the initial point to the terminal point. So, the magnitude of a vector is a positive number which is the measure of the line segment representing that vector. Magnitude of a vector is also known as the modulus or the absolute value of the vector.

Vectors and its Components

Vectors and its Components

Vectors: Introduction

The quantities which can be measured are called physical quantities. Some physical quantities have magnitudes only but some physical quantities have magnitudes as well as directions.

Let us study the addition of some physical quantities. For example, if we have two rectangles of areas 6cm2 and 8cm2, the total areas of these two rectangles is (6 + 8)cm2 i.e. 14cm2.

Sum of areas of two rectangles of area 6cm2 and  8cm2 is equal to 14cm2.

If two forces 4N and 5N act on a body, what is the total force acting on the body, 9N or 1N? Its answer is uncertain unless their directions are known. If two forces have same direction the total force on the body is 4N + 5N = 9N.

Sum (or resultant force) of two forces 4N and 5N acting toward the same direction is 9N i.e. 4N + 5N = 9N.

If two forces have opposite direction the total force on the body is 5N – 4N = 1N.

Sum (or resultant force) of two forces 4N and 5N acting toward the opposite direction is 1N i.e. 5N + (-4N) = 1N.

In the calculation of some physical quantities, the direction has very important role.

Solving Equations by Matrix Method

Solving Equations by Matrix Method

Solving Equations by Matrix Method

The simultaneous equations in two variables can be solved by different methods: substitution method, cross multiplication method etc. Here, we deal with the method of solving linear equations of two variables using matrices.

Let us consider the following linear equations,

a1x + b1y = c1 ………………. (i)

a2x + b2y = c2 ………………. (ii)

Writing equations (i) and (ii) in the matrix form, we get

(■(a_1&b_1@a_2&b_2 ))(■(x@y)) = (■(c_1@c_2 )) i.e.  AX = B ………………….. (iii) Where  A = (■(a_1&b_1@a_2&b_2 )),  X = (■(x@y))  and  B = (■(c_1@c_2 )) Pre-multiplying both sides of equation (iii) by inverse of A i.e. A-1, we get A-1(AX) = A-1B Or,  (A-1A)X = A-1B Or,  IX = A-1B   [∵ A-1A = I] ∴  X = A-1B   [∵  IX = X]

Equating the corresponding elements of equal matrices X and A-1B, we get the solution.

Note: If the determinant |A| = 0, the system of simultaneous equations have no unique solution, therefore when |A| ≠ 0, the equations have unique solution.

 

Worked Out Examples

Example 1: Solve uning matrices: x – 2y = -7 and 3x + 7y = 5 Solution: Here, x – 2y = -7 …………. (i) 3x + 7y = 5 ………… (ii) Writing equations (i) and (ii) in matrix form, we get (■(1&-2@3&7))(■(x@y)) = (■(-7@5)) i.e. AX = B Where, A = (■(1&-2@3&7)), X = (■(x@y)) and B = (■(-7@5)) ∴ |A| = |■(1&-2@3&7)| = 7 + 6 = 13 ∴ A-1 = 1/(|A|) (■(d&-b@-c&a)) = 1/13 (■(7&2@-3&1)) = (■(7/13&2/13@(-3)/13&1/13)) Now, by using formula X = A-1B, we have X = A-1B i.e.  (■(x@y)) = (■(7/13&2/13@(-3)/13&1/13))(■(-7@5)) Or,  (■(x@y)) = (■(7/13×-7+2/13×5@(-3)/13×-7+1/13×5)) Or,  (■(x@y)) = (■(-3@2)) Comparing the corresponding elements, x = -3, y = 2


Example 2: Solve by matrix method  4/x + 3/y = 7 and  3/x + 2/y = 4. Solution: Here, 4/x + 3/y = 7 ………………….. (i) 3/x + 2/y = 4 …………………. (ii) Writing equations (i) and (ii) in matrix form, we get (■(4&3@3&2))(■(1/x@1/y)) = (■(7@4)) i.e.  AX = B Where, A = (■(4&3@3&2)), X = (■(1/x@1/y)) and B = (■(7@4)) ∴|A| = |■(4&3@3&2)| = 8 – 9 = -1 ∴ A-1 = 1/(|A|) (■(d&-b@-c&a)) = 1/(-1) (■(2&-3@-3&4)) = (■(-2&3@3&-4)) Now, by using formula X = A-1B, we have X = A-1B i.e. (■(1/x@1/y)) = (■(-2&3@3&-4))(■(7@4)) = (■(-14+12@21-16)) = (■(-2@5)) Comparing the corresponding elements, 1/x = -2  or, x = (-1)/2   1/y = 5  or, y = 1/5 ∴  The values of x and y are (-1)/2  and 1/5.

Example 3: If the cost of 17 kg sugar and 4 kg tea is Rs. 1110 and the cost of 8 kg sugar and 2 kg tea is Rs. 540. Find the cost of sugar per kg and tea per kg by matrix method.

Solution: Let,

Cost of sugar per kg = Rs. x

Cost of tea per kg = Rs. y

Then according to questions,

1st case, 17x + 4y = 1110 …………… (i)

2nd case, 8x + 2y = 540 ……………… (ii)

Writing the given equations in the matrix form, we get

(■(17&4@8&2))(■(x@y)) = (■(1110@540)) i.e. AX = B Where, A = (■(17&4@8&2)), X = (■(x@y)) and B = (■(1110@540)) ∴ |A| = |■(17&4@8&2)| = 34 – 32 = 2 ≠ 0 So, the given equations have unique solution. Now, A-1 = 1/(|A|) (■(d&-b@-c&a)) = 1/2 (■(2&-4@-8&17))  Now, X = A-1B i.e.  (■(x@y)) = 1/2 (■(2&-4@-8&17))(■(1110@540)) Or,  (■(x@y)) = 1/2 (■(2×1110-4×540@-8×1110+17×540)) Or,  (■(x@y)) = 1/2 (■(60@300)) = (■(30@150))

Comparing the corresponding elements, we get

x = 30, y = 150

Hence, The cost of sugar per kg = Rs. 30

           The cost of tea per kg = Rs. 150

 

You can comment your questions or problems regarding the solution of simultaneous equations by matrix method here.

Minors and Cofactors of a Matrix

Minors and Cofactors of a Matrix

Minors and Cofactors of a Matrix

Minors:

Let A be a square matrix and aij is the element in ith row and jth column of A. Then the minor of the element aij is the determinant of the matrix formed by omitting ith row and jth column of A. The minor of element aij is denoted by Mij.

Adjoint of a Matrix

Adjoint of a Matrix

Adjoint of a Matrix

Let A be a square matrix, and Aij be the cofactors of the elements aij of the matrix A, then adjoint or adjugate of A denoted by adj A is the matrix obtained by transposing the matrix of cofactors of A.