Question

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

Answer 1

$\because$ The given vectors are co-planar

For co-planar Vectors,

${\overrightarrow{r}}_1.({\overrightarrow{r}}_2\times {\overrightarrow{r}}_3)=0$

$\Rightarrow \left|\begin{array}{ccc}a& 1& 1\\ 1& b& 1\\ 1& 1& c\end{array}\right|=0$

$\Rightarrow a(bc-1)-1(c-1)+1(1-b)=0\Rightarrow abc-a-c+1+1-b=0\Rightarrow abc+2=a+b+c\to \left(1\right)$

$\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}=\frac{(1-c-b+bc)+(1-c-a+ac)+(1-b-a+ab)}{(1-b-a+ab)(1-c)}$

$\Rightarrow =\frac{3-2(a+b+c)+(ab+bc+ac)}{1-c-b+bc-a+ac+ab-abc}$

$\Rightarrow =\frac{3-2(a+b+c)+(ab+bc+ac)}{1-(a+b+c)+(ab+bc+ac)-abc}$

Substituting $a+b+c=abc+2\to from\left(1\right)$

$\Rightarrow =\frac{3-2(abc+2)+(ab+bc+ac)}{1-2abc-2+ab+bc+ac}$

$\Rightarrow =\frac{3-2abc-4+(ab+bc+ac)}{-1-2abc+ab+bc+ac}$

$\Rightarrow =\frac{-1-2abc+(ab+bc+ac)}{-1-2abc+(ab+bc+ac)}$

$\Rightarrow =1$