question_answer

If the vector \[\vec{\alpha }=a\hat{i}+\hat{j}+\hat{k},\] \[\vec{\beta }=\hat{i}+b\hat{j}+\hat{k}\] and \[\vec{\gamma }=\hat{i}+\hat{j}+c\hat{k}\] are coplanar, then prove that. \[\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}=1,\] where \[\alpha \ne 1,\] \[b\ne 1\] \[c\ne 1.\]

Answer:

It is given that \[\vec{\alpha },\] \[\vec{\beta }\] and \[\vec{\gamma }\] are coplanar vectors. \[\therefore \]    \[[\vec{\alpha }\,\,\vec{\beta }\,\,\vec{\gamma }]=0\] \[\Rightarrow \]   \[\left| \begin{matrix}    a & 1 & 1  \\    1 & b & 1  \\    1 & 1 & c  \\ \end{matrix} \right|=0\] \[\Rightarrow \]   \[abc-a-c+1+1-b=0\] \[\Rightarrow \]   \[abc=a+b+c-2\]                         ?. (i) Now, \[\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}\] \[=\frac{(1-b)(1-c)+(1-c)(1-a)+(1-a)(1-b)}{(1-a)(1-b)(1-c)}\] \[=\frac{3-2(a\,+b+c)+(ab\,+bc\,+ca)}{1-(a\,+b\,+c)+(ab\,+bc\,+ca)-abc}\] \[=\frac{3-2(a\,+b\,+c)+ab\,+bc\,+ca}{1-(a\,+b\,+c)+(ab\,+bc\,+ca)-(a\,+b\,+c-2)}\]                                                 [Using Eq. (i)] \[=\frac{3-2(a\,+b\,+c)+ab\,+bc\,+ca}{3-2(a\,+b\,+c)+ab\,+bc\,+ca}=1\] Hence proved.