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*}
• 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*}
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