ONE MORE STEP TOWARD BETTER TOMORROW : CHAPTER 23 -Worked Out Examples

Friday, 8 August 2014

CHAPTER 23 -Worked Out Examples – 2

Example: 30

If the vector

	$\vec \alpha = a\hat i + \hat j + \hat k \,\,\,\, ,a \ne 1$
	$\vec \beta = \hat i + b\hat j + \hat k \,\,\,\, ,b \ne 1$
	$\vec \gamma = \hat i + \hat j + c\hat k \,\,\,\, ,c \ne 1$

are coplanar, prove that

$\dfrac{1}{{1 - a}} + \dfrac{1}{{1 - b}} + \dfrac{1}{{1 - c}} = 1$

Solution: 30

The coplanarity of the three vectors means that their $STP$ must be zero:

	$\Rightarrow \,\,\,\, \left\| {\;\begin{array}{*{20}{c}} a&1&1\\ 1&b&1\\ 1&1&c \end{array}\;} \right\|\; = 0$
	$\Rightarrow \,\,\,\, a(bc - 1) + (1 - c) + (1 - b) = 0$
	$\Rightarrow \,\,\,\, a + b + c = abc + 2$	$\ldots(1)$

We now have

	$\dfrac{1}{{1 - a}} + \dfrac{1}{{1 - b}} + \dfrac{1}{{1 - c}}$
	$= \dfrac{{(1 - b)(1 - c) + (1 - a)(1 - c) + (1 - a)(1 - b)}}{{(1 - a)(1 - b)(1 - c)}}$
	$= \dfrac{{3 - 2(a + b + c) + (ab + bc + ac)}}{{1 - (a + b + c) + (ab + bc + ac) - abc}}$
	$= \dfrac{{3 - 2(a + b + c) + (ab + bc + ac)}}{{1 - (a + b + c) + (ab + bc + ac) + 2 - (a + b + c)}}$ Using $(1)$ for the value of $abc$
	$=1$

Example: 31

Let $\vec a,\;\,\vec b,\;\,\vec c$ be three non-zero vectors such that $\vec c$ is a unit vector perpendicular to both $\vec a$ and $\vec b$ . If the angle between $\vec a$ and $\vec b$ is $\dfrac{\pi }{6}$ , prove that

${[\vec a\;\;\vec b\;\;\vec c]^2} = \dfrac{1}{4}{\left| {\vec a} \right|^2}\;\;{\left| {\vec b} \right|^2}$

Solution: 31

Since $\vec c$ is perpendicular to both $\vec a$ and $\vec b,\,\,\vec c$ must be parallel to $\vec a\;\; \times \;\vec b,$ i.e, the angle between $\vec c$ and $(\vec a \times \vec b)$ must be $0$ . Thus,

	$[\vec a\;\,\vec b\;\,\vec c] = [\vec c\;\,\vec a\;\,\vec b]$
	$= \vec c \cdot (\vec a \times \vec b)$
	$= \left\| {\vec c} \right\|\;\left\| {\vec a \times \vec b} \right\|\cos 0$
	$= 1 \cdot \left\| {\vec a} \right\|\;\left\| {\vec b} \right\|\sin \dfrac{\pi }{6} \cdot 1$
	$= \dfrac{1}{2}\left\| {\vec a} \right\|\;\left\| {\vec b} \right\|$	$\ldots(1)$

Squaring both sides of $(1)$ proves the stated assertion.

Example: 32

For three arbitrary vectors $\vec a,\;\;\vec b,\;\,\vec c,$ prove that

${[\vec a\;\;\vec b\;\vec c]^2} = \left| {\begin{array}{*{20}{c}} {\vec a \cdot \vec a}&{\vec a \cdot \vec b}&{\vec a \cdot \vec c}\\ {\vec b \cdot \vec a}&{\vec b \cdot \vec b}&{\vec b \cdot \vec c}\\ {\vec c \cdot \vec a}&{\vec c \cdot \vec b}&{\vec c \cdot \vec c} \end{array}} \right|$

Solution: 32

The relation is most easily proved by assuming $\vec a,\,\,\vec b,\,\,\vec c$ in rectangular form:

	$\vec a = {a_1}\hat i + {a_2}\hat j + {a_3}\hat k$
	$\vec b = {b_1}\hat i + {b_2}\hat j + {b_3}\hat k$
	$\vec c = {c_1}\hat i + {c_2}\hat j + {c_3}\hat k$

There’s no loss of generality in this assumption.

Now,

	${[\vec a\;\;\vec b\;\;\vec c]^2} = {\left\| {\;\begin{array}{*{20}{c}} {{a_1}}&{{a_2}}&{{a_3}}\\ {{b_1}}&{{b_2}}&{{b_3}}\\ {{c_1}}&{{c_2}}&{{c_3}} \end{array}\;} \right\|^2}$
	$= \left\| {\;\begin{array}{{20}{c}} {{a_1}}&{{a_2}}&{{a_3}}\\ {{b_1}}&{{b_2}}&{{b_3}}\\ {{c_1}}&{{c_2}}&{{c_3}} \end{array}\;} \right\|\;\left\| {\;\begin{array}{{20}{c}} {{a_1}}&{{a_2}}&{{a_3}}\\ {{b_1}}&{{b_2}}&{{b_3}}\\ {{c_1}}&{{c_2}}&{{c_3}} \end{array}\;} \right\|$
	$= \left\| {\begin{array}{*{20}{c}} {a_1^2 + a_2^2 + a_3^2}&{{a_1}{b_1} + {a_2}{b_2} + {a_3}{b_3}}&{{a_1}{c_1} + {a_2}{c_2} + {a_3}{c_3}}\\ {{b_1}{a_1} + {b_2}{a_2} + {b_3}{a_3}}&{b_1^2 + b_2^2 + b_3^2}&{{b_1}{c_1} + {b_2}{c_2} + {b_3}{c_3}}\\ {{c_1}{a_1} + {c_2}{a_2} + {c_3}{a_3}}&{{c_1}{b_1} + {c_2}{b_2} + {c_3}{b_3}}&{c_1^2 + c_2^2 + c_3^2} \end{array}} \right\|$
	$= \left\| {\,\begin{array}{*{20}{c}} {\vec a \cdot \vec a}&{\vec a \cdot \vec b}&{\vec a \cdot \vec c}\\ {\vec b \cdot \vec a}&{\vec b \cdot \vec b}&{\vec b \cdot \vec c}\\ {\vec c \cdot \vec a}&{\vec c \cdot \vec b}&{\vec c \cdot \vec c} \end{array}\,} \right\|$

Friday, 8 August 2014

CHAPTER 23 -Worked Out Examples – 2

No comments:

https://www.youtube.com/TarunGehlot