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Vectors

  • In mathematics, a vector is an element of a vector space.
  • A vector has both magnitude and direction.
  • It is commonly represented using a letter with an arrow on top, such as [ \vec{A} ]. \

Vector Addition

  • Vectors can be added together to create a resultant vector.
  • The addition is done by adding the corresponding components of the vectors.
  • For example, for two vectors [ \vec{A} = (a_1, a_2) ] and [ \vec{B} = (b_1, b_2) ], the resultant vector [ \vec{R} ] is obtained by [ \vec{R} = \vec{A} + \vec{B} = (a_1 + b_1, a_2 + b_2) ]. \

Scalar Multiplication

  • A vector can be multiplied by a scalar (a real number) to change its magnitude.
  • The scalar multiplication is done by multiplying each component of the vector by the scalar.
  • For example, if [ \vec{A} = (a_1, a_2) ] and [ k ] is a scalar, then [ k\vec{A} = (k \cdot a_1, k \cdot a_2) ]. \

Dot Product

  • The dot product (also known as scalar product) of two vectors is a scalar quantity.
  • It is calculated by multiplying the corresponding components of the vectors and then taking the sum.
  • The dot product of two vectors [ \vec{A} = (a_1, a_2) ] and [ \vec{B} = (b_1, b_2) ] is given by [ \vec{A} \cdot \vec{B} = a_1 \cdot b_1 + a_2 \cdot b_2 ]. \

Cross Product

  • The cross product (also known as vector product) of two vectors is a vector quantity.
  • It is calculated using a determinant formula.
  • The cross product of two vectors [ \vec{A} = (a_1, a_2, a_3) ] and [ \vec{B} = (b_1, b_2, b_3) ] is given by: [ \vec{A} \times \vec{B} = (a_2 \cdot b_3 - a_3 \cdot b_2, a_3 \cdot b_1 - a_1 \cdot b_3, a_1 \cdot b_2 - a_2 \cdot b_1) ]. \

Vector Magnitude

  • The magnitude (length or size) of a vector [ \vec{A} ] is denoted by [ |\vec{A}| ].
  • It can be calculated using the Pythagorean theorem.
  • For a vector [ \vec{A} = (a_1, a_2) ], the magnitude [ |\vec{A}| ] is given by [ |\vec{A}| = \sqrt{a_1^2 + a_2^2} ]. \

Unit Vector

  • A unit vector is a vector with magnitude equal to 1.
  • It is commonly used to represent direction.
  • The unit vector in the direction of a vector [ \vec{A} ] is denoted by [ \hat{A} ].
  • It can be calculated by dividing each component of [ \vec{A} ] by its magnitude: [ \hat{A} = \frac{\vec{A}}{|\vec{A}|} ]. \

Vector Components

  • A vector can be broken down into its components along different axes.
  • For a vector [ \vec{A} ], its components are usually denoted by [ A_x ] and [ A_y ] in a 2D coordinate system.
  • The components can be calculated using the magnitudes and direction cosine ratios of the vector. \

Magnitude of Vector Components

  • The magnitude of a vector [ \vec{A} ] in terms of its components in a 2D coordinate system can be calculated using the Pythagorean theorem.
  • For a vector [ \vec{A} = (A_x, A_y) ], its magnitude [ |\vec{A}| ] is given by [ |\vec{A}| = \sqrt{A_x^2 + A_y^2} ]. \

Example

Given vectors [ \vec{A} = 2\hat{i} + 3\hat{j} ] and [ \vec{B} = 4\hat{i} - 5\hat{j} ], find:

  1. The sum of vectors [ \vec{A} ] and [ \vec{B} ].
  1. The dot product of vectors [ \vec{A} ] and [ \vec{B} ]. [ \vec{A} + \vec{B} = (2+4)\hat{i} + (3-5)\hat{j} = 6\hat{i} - 2\hat{j} ] [ \vec{A} \cdot \vec{B} = (2 \cdot 4) + (3 \cdot -5) = -7 ] \

Vectors - Problems

  • Vectors can be used to solve various problems in mathematics and physics.
  • Some common problems involving vectors include finding the displacement, velocity, and acceleration of an object.
  • Vectors are also used in trigonometry to solve problems related to forces, angles, and distances. \

Proving Triangle is Isosceles

  • To prove that a triangle is isosceles, we can use vectors.
  • If we have three vectors representing the sides of a triangle, and two of those vectors are equal in magnitude and direction, then the triangle is isosceles.
  • We can use the properties of vector addition and the properties of equal vectors to prove this. \

Example 1

Given vectors [ \vec{A} = 3\hat{i} + 2\hat{j} ] and [ \vec{B} = -2\hat{i} + 3\hat{j} ], prove that triangle formed by [ \vec{A} ], [ \vec{B} ], and the origin [ \vec{O} ] is isosceles. Step 1: Calculate the vectors [ \vec{AB} ] and [ \vec{AO} ] using vector subtraction. [ \vec{AB} = \vec{B} - \vec{A} = (-2\hat{i} + 3\hat{j}) - (3\hat{i} + 2\hat{j}) = -5\hat{i} + \hat{j} ] [ \vec{AO} = \vec{O} - \vec{A} = (0\hat{i} + 0\hat{j}) - (3\hat{i} + 2\hat{j}) = -3\hat{i} - 2\hat{j} ] \

Example 1 (continued)

Step 2: Calculate the magnitudes of vectors [ \vec{AB} ] and [ \vec{AO} ] using the magnitude formula. [ |\vec{AB}| = \sqrt{(-5)^2 + 1^2} = \sqrt{26} ] [ |\vec{AO}| = \sqrt{(-3)^2 + (-2)^2} = \sqrt{13} ] Step 3: Compare the magnitudes of [ \vec{AB} ] and [ \vec{AO} ]. Since [ |\vec{AB}| = \sqrt{26} ] and [ |\vec{AO}| = \sqrt{13} ], and [ \sqrt{26} ] is not equal to [ \sqrt{13} ], the triangle is not isosceles. \

Example 2

Given vectors [ \vec{CD} = 4\hat{i} + 6\hat{j} ] and [ \vec{DE} = 2\hat{i} - 3\hat{j} ], prove that triangle formed by [ \vec{C} ], [ \vec{D} ], and [ \vec{E} ] is isosceles. Step 1: Calculate the vectors [ \vec{CE} ] and [ \vec{CD} ] using vector subtraction. [ \vec{CE} = \vec{E} - \vec{C} = (2\hat{i} - 3\hat{j}) - (4\hat{i} + 6\hat{j}) = -2\hat{i} - 9\hat{j} ] [ \vec{CD} = \vec{D} - \vec{C} = (4\hat{i} + 6\hat{j}) - (0\hat{i} + 0\hat{j}) = 4\hat{i} + 6\hat{j} ] \

Example 2 (continued)

Step 2: Calculate the magnitudes of vectors [ \vec{CE} ] and [ \vec{CD} ] using the magnitude formula. [ |\vec{CE}| = \sqrt{(-2)^2 + (-9)^2} = \sqrt{85} ] [ |\vec{CD}| = \sqrt{(4)^2 + (6)^2} = \sqrt{52} ] Step 3: Compare the magnitudes of [ \vec{CE} ] and [ \vec{CD} ]. Since [ |\vec{CE}| = \sqrt{85} ] and [ |\vec{CD}| = \sqrt{52} ], and [ \sqrt{85} ] is not equal to [ \sqrt{52} ], the triangle is not isosceles. \

Applications of Vectors

  • Vectors are widely used in physics to represent quantities such as displacement, velocity, and acceleration.
  • They are used in engineering to represent forces, moments, and electric fields.
  • Vectors are also used in computer graphics to represent positions, directions, and transformations.
  • In mathematics, vectors are used in linear algebra, calculus, and geometry. \

Vector Algebra

  • Vector algebra is the branch of algebra that deals with vector operations such as addition, subtraction, and multiplication.
  • Vector algebra is used to solve equations involving vectors, derive results, and prove theorems.
  • It provides a systematic way to manipulate vectors and analyze their properties. \

Vector Calculus

  • Vector calculus is the branch of calculus that deals with differentiation and integration of vector functions.
  • It extends the concepts of differentiation and integration to vector-valued functions.
  • Vector calculus is used in physics, engineering, and mathematics to model and analyze various phenomena, such as fluid flow, electromagnetic fields, and gradients. \

Summary

  • Vectors have magnitude and direction, and are represented using letters with arrows on top.
  • Vector addition is done by adding corresponding components, and scalar multiplication is done by multiplying each component by a scalar.
  • The dot product of two vectors is a scalar, while the cross product is a vector.
  • Vectors can be broken down into components and their magnitudes calculated.
  • Vectors are used to solve problems, prove triangles are isosceles, and for various applications in physics, engineering, and mathematics.

Vector Equations

  • Vector equations represent relationships between vectors.
  • They can be written using variables that represent vectors.
  • For example, the equation [ \vec{A} + \vec{B} = \vec{C} ] shows that the sum of vectors [ \vec{A} ] and [ \vec{B} ] is equal to vector [ \vec{C} ].

Vector Projections

  • The projection of a vector [ \vec{A} ] onto another vector [ \vec{B} ] is a vector that represents the component of [ \vec{A} ] in the direction of [ \vec{B} ].
  • It can be calculated using the dot product and the magnitude of [ \vec{B} ].
  • The projection of [ \vec{A} ] onto [ \vec{B} ] is given by [ \text{proj}_{\vec{B}} \vec{A} = \left(\frac{\vec{A} \cdot \vec{B}}{|\vec{B}|^2}\right)\vec{B} ].

Example

Given vector [ \vec{A} = 3\hat{i} + 4\hat{j} ] and vector [ \vec{B} = \hat{i} - 2\hat{j} ], calculate the projection of [ \vec{A} ] onto [ \vec{B} ]. Step 1: Calculate the dot product of [ \vec{A} ] and [ \vec{B} ]. [ \vec{A} \cdot \vec{B} = (3 \cdot 1) + (4 \cdot -2) = -5 ] Step 2: Calculate the magnitude of [ \vec{B} ]. [ |\vec{B}| = \sqrt{1^2 + (-2)^2} = \sqrt{5} ] Step 3: Calculate the projection of [ \vec{A} ] onto [ \vec{B} ]. [ \text{proj}_{\vec{B}} \vec{A} = \left(\frac{\vec{A} \cdot \vec{B}}{|\vec{B}|^2}\right)\vec{B} = \left(\frac{-5}{5}\right)(\hat{i} - 2\hat{j}) = -\hat{i} + 2\hat{j} ]

Vector Triple Product

  • The vector triple product is the product of three vectors.
  • It is calculated using the cross product and dot product.
  • For three vectors [ \vec{A} ], [ \vec{B} ], and [ \vec{C} ], the vector triple product is given by [ \vec{A} \times (\vec{B} \times \vec{C}) ].
  • It can also be expanded as [ (\vec{A} \cdot \vec{C})\vec{B} - (\vec{A} \cdot \vec{B})\vec{C} ].

Example

Given vectors [ \vec{A} = \hat{i} + 2\hat{j} + \hat{k} ], [ \vec{B} = 2\hat{i} - \hat{j} ], and [ \vec{C} = -\hat{j} + 3\hat{k} ], calculate the vector triple product [ \vec{A} \times (\vec{B} \times \vec{C}) ]. Step 1: Calculate the cross product of [ \vec{B} ] and [ \vec{C} ]. [ \vec{B} \times \vec{C} = (2\hat{i} - \hat{j}) \times (-\hat{j} + 3\hat{k}) ]
[ = (-\hat{j} + 7\hat{k}) ]
Step 2: Calculate the cross product of [ \vec{A} ] and [ (\vec{B} \times \vec{C}) ]. [ \vec{A} \times (\vec{B} \times \vec{C}) = (\hat{i} + 2\hat{j} + \hat{k}) \times (-\hat{j} + 7\hat{k}) ]
[ = -5\hat{i} - \hat{j} + 15\hat{k} ] \

Vector Triple Product - Properties

  • The vector triple product satisfies the following properties:
    • [ \vec{A} \times (\vec{B} \times \vec{C}) = (\vec{A} \cdot \vec{C})\vec{B} - (\vec{A} \cdot \vec{B})\vec{C} ] (expanded form)
    • [ \vec{A} \times (\vec{B} \times \vec{C}) = \vec{B} \times (\vec{C} \times \vec{A}) = \vec{C} \times (\vec{A} \times \vec{B}) ] (cyclic property)
    • [ \vec{A} \times (\vec{B} \times \vec{C}) + \vec{B} \times (\vec{C} \times \vec{A}) + \vec{C} \times (\vec{A} \times \vec{B}) = \vec{0} ] (scalar triple product form)

Vector Differentiation

  • Vector differentiation deals with finding the derivatives of vector functions.
  • It extends the concepts of differentiation (limit, derivative, tangent) to vector-valued functions.
  • Key rules for vector differentiation include