JEE Main & Advanced
Mathematics
Vector Algebra
Question Bank
Critical Thinking
question_answer
If the vectors \[a\mathbf{i}+\mathbf{j}+\mathbf{k},\,\,\mathbf{i}+b\mathbf{j}+\mathbf{k}\] and \[\mathbf{i}+\mathbf{j}+c\mathbf{k}\] \[(a\ne b\ne c\ne 1)\] are coplanar, then the value of \[\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}=\] [BIT Ranchi 1988; RPET 1987; IIT 1987; DCE 2001; MP PET 2004; Orissa JEE 2005]
A)- 1
B)\[-\frac{1}{2}\]
C)\[\frac{1}{2}\]
D)1
Correct Answer:
D
Solution :
Since \[\left| \begin{matrix} a & 1 & 1 \\ 1 & b & 1 \\ 1 & 1 & c \\ \end{matrix} \right|=0\]
Applying \[{{R}_{2}}\to {{R}_{2}}-{{R}_{1}}\] and \[{{R}_{3}}\to {{R}_{3}}-{{R}_{1}},\]
we get \[\left| \begin{matrix} a & 1 & 1 \\ 1-a & b-1 & 0 \\ 1-a & 0 & c-1 \\ \end{matrix} \right|=0\]
On expanding, we get
\[a(b-1)(c-1)-(1-a)(c-1)-(1-a)(b-1)=0\]
On dividing by \[(1-a)(1-b)(1-c),\] we get \[\frac{a}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}=0\]