Question

If the vectors $a\hat{i}+\hat{j}+\hat{k},\hat{i}+b\hat{j}+\hat{k},\hat{i}+\hat{j}+c\hat{k}(a\ne 1,b\ne 1,c\ne 1)$ are coplanar, then the value of $\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}$

Answer 1

Given: $\left|\begin{array}{ccc}a& 1& 1\\ 1& b& 1\\ 1& 1& c\end{array}\right|$ are coplanar
$\Rightarrow \left|\begin{array}{ccc}a& 1& 1\\ 1& b& 1\\ 1& 1& c\end{array}\right|=0$
$R_2\to R_2-R_1,R_3\to R_3-R_1$
$\Rightarrow \left|\begin{array}{ccc}a& 1& 1\\ 1-a& b-1& 0\\ 1-a& 0& c-1\end{array}\right|=0$
Divide $C_1$,by $(1-a)$, $C_2$ by $(1-b)$,and $C_3$,by $(1-c)$
$\Rightarrow \left|\begin{array}{ccc}\frac{a}{1-a}& \frac{-1}{1-b}& \frac{-1}{1-c}\\ 1& 1& 0\\ 1& 0& 1\end{array}\right|=0$
$\Rightarrow \frac{a}{1-a}(1-0)-\frac{-1}{1-b}(1-0)+\left(\frac{-1}{1-c}\right)(0-1)=0$
$\Rightarrow \frac{a}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}=0$
Add $1$ to both sides, we get
$\Rightarrow \frac{a}{1-a}+1+\frac{1}{1-b}+\frac{1}{1-c}=1$
$\Rightarrow \frac{a+1-a}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}=1$
$\Rightarrow \frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}=1$