Lecture 5 MATH 127 (Maj Sultana)

 

MATH 127 

Vector Analysis

    1) Prove that,  $|\vec{A} .\vec{B} |^2 + |\vec{A} \times \vec{B} |^2 = |\vec{A} |^2 .|\vec{B} |^2$  

Solution:  $$\begin{align*} |\vec{A} . \vec{B} |^2 & = |\vec{A} |^2 + |\vec{B} |^2 \cos^2 \theta \\ |\vec{A} \times \vec{B} |^2 & = |\vec{A} |^2. |\vec{B} |^2 \sin^2 \theta |\hat{n} |^2 \\ & = |\vec{A} |^2 . |\vec{B} |^2 \sin^2 \theta \\ \therefore \textrm{ L.S} & = |\vec{A} |^2 . |\vec{B} |^2 ( \sin^2 \theta + \cos^2 \theta ) \\ & = |\vec{A} |^2 | \vec{B} |^2 \\ & = \textrm{ R.S} \end{align*}$$  

    2) Prove that,  $\vec{A} .( \vec{B} \times \vec{C} ) = \vec{B} .(\vec{C} \times \vec{A} ) = \vec{C} . ( \vec{A} \times \vec{B} )$  

Solution: [Hint:   Let,  $$\begin{align*} & \vec{A} = A_1 \hat{i} + A_2 \hat{j} + A_3 \hat{k} \\ & \vec{B} = B_1 \hat{i} + B_2 \hat{j} + B_3 \hat{k} \\ & \vec{C} = C_1 \hat{i} + C_2 \hat{j} + C_3 \hat{k} \end{align*}$$  

   ♦ Reciprocal sets of vectors: 

the set $a, b, c$ and $a', b',c'$ are called reciprocal sets of reciprocal system of vectors if  $$\begin{align*} & a.a' = b.b' = c.c' = 1 \\ \implies & ~~~~~~~~~~~ \boxed {a = \frac{1}{a'}} \\ \textrm{and, } & a'b = a'c = b'a = b'c = c'a = c'b = 0 \end{align*}$$                                                        

    That is, each vector is orthogonal to the reciprocal of the other two vectors in the system.  $\boxed {a = \frac{1}{a'}}$  

    

   ♦ Theorem: The sets $a,b,c$ and $a',b',c'$ are reciprocal sets of vectors if and only if  $$\begin{align*} & a' = \frac{b \times c}{a.b\times c} \\ & b' = \frac{c \times a}{a.b\times c} ~~~~~~~~~\big| \textrm{where,} a.b\times c \ne 0\\ & c' = \frac{a \times b}{a.b\times c} \\ & ~ \\ & \boxed{a.b\times c = b.c\times a = c.a\times b} \end{align*}$$  

    Example: Find a set of vectors reciprocal to the set vector. 

Solution: Let,  $$\begin{align*} \vec{a} & = 2 \hat{i} + 3 \hat{j} - \hat{k} \\ \vec{b} & = \hat{i} - \hat{j} - 2 \hat{k} \\ \vec{c} & = - \hat{i} + 2 \hat{j} + 2 \hat{k} \\ & ~~~ \\ \textrm{Now,}& \\ ( \vec{b} \times \vec{c} & = \left| \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & - 1 & - 2 \\ - 1 & 2 & 2 \end{array}) \right| \\ & = 2 \hat{i} + \hat{k} \\ & ~~ \\ & ~~ \\ \vec{a} .(\vec{b} \times \vec{c}) & = 4 - 1 = 3 \\ & ~~ \\ & ~~ \\ (\vec{c} \times \vec{a} ) & = \left| \begin{array}[]{ccc} \hat{i} & \hat{j} & \hat{k} \\ - 1 & 2 & 2 \\ 2 & 3 & - 1 \end{array} \right| \\ & = - 8 \hat{i} + 3 \hat{j} - 7 \hat{k} \\ & ~~ \\ (\vec{a} \times \vec{b} ) & = \left| \begin{array}[]{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & - 1 \\ 1 & - 1 & - 2 \end{array} \right| \\ & = - 7 \hat{i} + 3 \hat{j} - 5 \hat{k} \\ & ~ \\ & ~\\ \therefore a' & = \frac{b \times c }{a. b \times c} = \frac{2}{3} \hat{i} + \frac{1}{3}\hat{k} \\ b' & = \frac{c \times a}{a.b\times c} = - \frac{8}{3} \hat{i} + \hat{j} - \frac{7}{3} \hat{k} \\ c' & = \frac{a\times b}{a.b\times c} = - \frac{7}{3} \hat{i} + \hat{j} - \frac{5}{3} \hat{k} \end{align*}$$  

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    • Direction Cosine:  $$\cos \alpha = \frac{x}{|\vec{r}|} = \frac{x}{\sqrt{ x^2 + y^2 + z^2}}$$ $$\cos \beta = \frac{y}{|\vec{r}|} = \frac{y}{\sqrt{x^2 + y^2 + z^2}}$$ $$\cos \gamma = \frac{z}{|\vec{r}|} = \frac{z }{\sqrt{x^2 + y^2 + z^2} }$$  

    • Direction Ratio:  $$\begin{align*} \vec{r} = & x \hat{i} + y \hat{j} + z \hat{k} \\ & x: y: z \\ & \hat{l} = \frac{x}{\vec{r}} \\ & \hat{m} = \frac{y}{\vec{r}} \\ & \hat{n} = \frac{z}{\vec{r}} & ~ \\ & \boxed { l^2 + m^2 + n^2 = 1 } \end{align*}$$  

    ♦ Vector linear dependency and linear independence: 

Vectors  $\vec{A}_1, \vec{A}_2, \vec{A}_3, \ldots \ldots \vec{A}_n$ are linearly dependent if there exist scalers  $a_1, a_2, a_3 \ldots \ldots a_n$ not all zero, such that,

  $$\boxed{ \vec{A}_1 a_1 + \vec{A}_2 a_2 + \vec{A}_3 a_3 + \ldots \ldots \vec{A}_n a_n = 0} ~~ \rightarrow ~ \textrm{Linear Combination}$$  

Otherwise, the vectors are linearly independent.

    Example: Find the linear combination of the vectors  $\vec{a} = \hat{i} + \hat{j} - \hat{k}, ~ \vec{b} = \hat{i} - \hat{j} + \hat{k} \textrm{ and } \vec{c} = - \hat{i} + \hat{j} + \hat{k}$ . Test whether they are linearly dependent or not.

Solution: Let,  $x,y,z$ are scalers.

The linear combination is, 

  $$\begin{align*} & x \vec{a} + y \vec{b} + z \vec{c} \\ = & x (\hat{i} + \hat{j} - \hat{k}) + y (\hat{i} - \hat{j} + \hat{k} ) + z( - \hat{i} + \hat{j} + \hat{k} )\\ = & \hat{i} ( x + y + z) + \hat{j} (x - y + z) + \hat{k} (- x + y + z) \end{align*}$$  

This will be linearly dependent if,  $\hat{i} ( x + y + z) + \hat{j} (x - y + z) + \hat{k} (- x + y + z) = 0$  

  $$\begin{align*} \therefore & x + y - z = 0 \\ & z - y + z = 0 \\ & - x + y z = 0 & ~ \\ \therefore & x = y = z = 0 \end{align*}$$