Lecture 2 [06 Feb, 2020] MATH 127 (Maj Sultana)

 MATH 127 

Vector Analysis





    ♦ Vector addition: The sum of two vector \(\vec{A}\) and \(\vec{B}\) is a vector  \(\vec{C}\) , which obtained by placing the initial point of  \(\vec{B}\) on the final point of  \(\vec{A}\) and the drawing a line from the initial point of  \(\vec{A}\) to the final point of  \(\vec{B}\) .

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    ♦ Scaler product: The product of a scaler say m times a vector  \(\vec{A}\) is another vector  \(\vec{B}\) , where  \(\vec{B}\) has the direction as  \(\vec{A}\) but the magnitude is changed that is  \(|\vec{B}| = m |\vec{A}|\)  

    • Properties: 

1. Commutative law for addition:  \(\vec{A} + \vec{B} = \vec{B} + \vec{A}\)  

2. Associative law for addition:  \(\vec{A}+ (\vec{B} + \vec{C}) = (\vec{A} + \vec{B} ) + \vec{C}\)  

3. Commutative law for multiplication:  \(m \vec{A} = \vec{A}m \)  

4. Associative law for multiplication:  \((m + n) \vec{A} = m \vec{A} + n \vec{A} \) , where \(m\) and \(n\) are two different scalers.

5. Distributive law:  \(m(\vec{A} + \vec{B} ) = m\vec{A} + m \vec{B}\)  




    Vector Multiplication: There are two types of multiplication.

1. Scaler or Dot product          2. Vector or Cross product


    Dot Product: 

A geometric representation of dot product: The scaler or dot product of two vectors,  \(\vec{A}\)   and  \(\vec{B}\) denoted by  \(\vec{A}. \vec{B}\) , is defined as the product of the magnitude of vector times the cosine of the angle between them.

  \(\vec{A} . \vec{B} = |\vec{A}|~|\vec{B }| \cos \theta ~~~~~~~~[\theta = \textrm{mnimum angle} ]\)  


A cartesian representation of dot product:  \( textrm{I} \vec{A} = A_1 \hat{i} + A_2 \hat{j} + A_3 \hat{k} \textrm{ and } \vec{B} = B_1 \hat{i} + B_2 \hat{j} + B_3 \hat{k} \textrm{ then } \vec{A} \vec{B} = A_1 B_1 + A_2 B_2 + A_3 B_3 \)  


    • Properties of Dot product:

1.  \(\vec{A} .\vec{B} = \vec{B} .\vec{A} \)  

2.  \(\vec{A} .(\vec{B} + \vec{C} ) = \vec{A} . \vec{B} + \vec{A} .\vec{C} \)  

3.  \(m(\vec{A} .\vec{B} ) = (m \vec{A} .\vec{B} ) = \vec{A} .(m \vec{B} ) = (\vec{A} . \vec{B} ) m \)


    Some Important results:

1.  \(\hat{i} . \hat{i} = |\hat{i} | |\hat{i} | \cos 0 = 1 \) \(\hat{j} . \hat{j} = 1 \) \(\hat{k} . uvjk = 1 \)  

2.  \(\hat{i} . \hat{j} = |\hat{i} | . |\hat{j} | \cos 90 = 0 \) \(\hat{j} . \hat{k} = 0 \) \(\hat{k} . \hat{i} = 0 \)  

3. If we have  \(\vec{A} . \vec{B} = 0\) means, neither the magnitude of  \(\vec{A} \) nor  \(\vec{B} \) is 0, then  \(\vec{A} \) and  \(\vec{b}\) must be perpendicular.

  \begin{align*} & \vec{A} . \vec{B} = 0 and |\vec{A}| \ne 0 \textrm{ and } |\vec{B} | \ne 0 \\ & ~~~~~~~~~~~ \textrm{then, } \theta = 90^{\circ} \end{align*}  



    Example: What is the angle between of two vectors  \(\vec{A} = 2 \hat{i} + 2 \hat{j} \textrm{ and } \vec{B} = 4 \hat{i} - 3 \hat{j} \) ?

Solution:  \begin{align*} & \vec{A} . \vec{B} = 12 - 6 = 6\\ & |\vec{A} | = \sqrt{2^2 + 2^2} = 2 \sqrt{2}\\ & |\vec{B} | = \sqrt{6^2 + (- 3)^2} = \sqrt{45} \\ & ~~~ \therefore \theta = \cos^{-1} \left(\frac{6}{2\sqrt{90}}\right) \\ & ~~~ ~~~~~~~= 71.56^\circ \end{align*}




    ♦ Cross product: 

    A geometric representation: The cross product of two vector  \(\vec{A} \) and  \(\vec{B} \) denoted by  \(\vec{A} \times \vec{B} \) is defined as the product of the magnitudes of the vectors times the sine of the angle between them with a unit vector  \(\hat{n} \) .

  \(\vec{A} \times \vec{B} = |\vec{A} |.|\vec{B} | \sin \theta ~ \hat{n} \) is the unit vector at right angles to both \(\vec{A} \) and \(\vec{A} \)


    A cartesian representation: 

 If \begin{align*} & \vec{A} = A_1 \hat{i} + A_2 \hat{j} + A_3 \hat{k} \\ & \vec{B} = B_1 \hat{i} + B_2 \hat{j} + B_3 \hat{k} \end{align*}   

Then  \begin{align*} \vec{A} \times \vec{B} & = \left |\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ A_1 & A_2 & A_3 \\ B_1 & B_2 & B_3 \end{array} \right | \\ & = \hat{i} (A_2 B_3 - B_2 A_3 ) + \hat{j} (A_3 B_1 - A_1 B_3 ) + \hat{k} (A_1 B_2 - A_2 B_1) \end{align*}  



    Example: If  \(\vec{A} = 2 \hat{i} + \hat{j} - \hat{k} \textrm{ and } \vec{B} = - 3 \hat{i} + 4 \hat{j} + \hat{k} \) find  \(\vec{A} \times \vec{B} \)                        


Solution: 

  \begin{align*} \vec{A} \times \vec{B} = & \left| \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & - 1 \\ - 3 & 4 & 1 \end{array} \right| \\ = & 6 \hat{i} + \hat{j} + 11 \hat{k} \end{align*}  



    • Properties of Cross product: 

1. If  \(\vec{A} \times \vec{B} = 0 \) then  \(\vec{A} \) and  \(\vec{B} \) will be parallel vector.

2.  \(\vec{A} \times \vec{B} \ne 0\) provided that  \(\vec{A} \times \vec{B} \) is orthogonal to both  \(\vec{A} \) and  \(\vec{B} \)  

3. If  \(\vec{u}, \vec{v} \textrm{ and } \vec{w}\) vectors and c is a number then

  \begin{align*} i) &~ \vec{u} \times \vec{v} = - ( \vec{v} \times \vec{u} ) \\ ii) & ~\vec{u} \times (\vec{v} + \vec{w} ) = \vec{u} \times \vec{v} + \vec{u} \times \vec{w} \\ iii) & ~(c \vec{u} ) \times \vec{v} = uvu \times (c \vec{v} ) = c ( \vec{u} \times \vec{v} ) \\ iv) & ~ \vec{u} (\vec{v} \times \vec{w} )= (\vec{u} \times \vec{v} ) \vec{w} \\ v) & ~ \vec{u} ( \vec{v} \times \vec{w} ) = \left| \begin{array}{ccc} u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \\ w_1 & w_2 & w_3 \end{array} \right| \end{align*}  



  



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