A) 0
B) 1
C) -1
D) 2
Correct Answer: B
Solution :
Key Idea: If three vectors are coplanar, then the values of the determinant should be zero. Since, the vectors, \[a\hat{i}+\hat{j}+\hat{k},\hat{i}+b\hat{j}+\hat{k}\] and \[\hat{i}+\hat{j}+c\hat{k}\]are coplanar \[\therefore \] \[\left| \begin{matrix} a & 1 & 1 \\ 1 & b & 1 \\ 1 & 1 & c \\ \end{matrix} \right|=0\] Appling \[{{C}_{1}}\to {{C}_{1}}-{{C}_{3}},{{C}_{2}}\to {{C}_{2}}-{{C}_{3}}\] \[\Rightarrow \] \[\left| \begin{matrix} a-1 & 0 & 1 \\ 0 & b-1 & 1 \\ 1-c & 1-c & c \\ \end{matrix} \right|=0\] \[\Rightarrow \] \[(a-1)[(b-1)c-1(1-c)]\] \[\Rightarrow \] \[-1(b-1)(1-c)=0\] \[\Rightarrow \] \[(a-1)(bc-1)-(b-1)(1-c)=0\] \[\Rightarrow \]\[\frac{(bc-1)}{(b-1)(1-c)}-\frac{1}{(a-1)}=0\] \[\Rightarrow \]\[\frac{1}{(1-a)}+\frac{(1-c)+(1-b)c}{(1-b)(1-c)}=0\] \[\Rightarrow \]\[\frac{1}{(1-a)}+\frac{1}{(1-b)}+\frac{c}{(1-c)}+1=1\] \[\Rightarrow \]\[\frac{1}{(1-a)}+\frac{1}{(1-b)}+\frac{c+1-c}{(1-c)}=1\] \[\Rightarrow \]\[\frac{1}{(1-a)}+\frac{1}{(1-b)}+\frac{1}{(1-c)}=1\]
You need to login to perform this action.
You will be redirected in
3 sec
