Vector Algebra Question 175
Question: 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]
Options:
A) - 1
B) $ -\frac{1}{2} $
C) $ \frac{1}{2} $
D) 1
Show Answer
Answer:
Correct Answer: D
Solution:
- Since $ \begin{vmatrix} a & 1 & 1 \\ 1 & b & 1 \\ 1 & 1 & c \\ \end{vmatrix} =0 $
Applying $ R_2\to R_2-R_1 $ and $ R_3\to R_3-R_1, $
we get $ \begin{vmatrix} a & 1 & 1 \\ 1-a & b-1 & 0 \\ 1-a & 0 & c-1 \\ \end{vmatrix} =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 $
$ \Rightarrow \frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}=\frac{1}{1-a}-\frac{a}{1-a}=1. $
BETA
