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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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