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} ].
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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) ].
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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) ].
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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 ].
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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) ].
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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} ].
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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}|} ].
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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.
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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} ].
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Example
Given vectors [ \vec{A} = 2\hat{i} + 3\hat{j} ] and [ \vec{B} = 4\hat{i} - 5\hat{j} ], find:
- The sum of vectors [ \vec{A} ] and [ \vec{B} ].
- 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 ]
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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.
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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.
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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} ]
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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.
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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} ]
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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.
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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.
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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.
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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.
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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