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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|
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