Question

If the vectors $a\overline{i}+\overline{j}+\overline{k},\overline{i}+b\overline{j}+\overline{k},\overline{i}+\overline{j}+c\overline{k}$ ($a,b,c$ are not equal to $1$) are coplanar, then
$\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}=.....$

• A. $0$
• B. $3$
• C. $2$
• D. $1$

Answer 1

The correct option is D $1$

$\begin{array}{c}\left|\begin{array}{ccc}a& 1& 1\\ 1& b& 1\\ 1& 1& c\end{array}\right|=0\\ c_1\to c_1-c_2{c}_2\to c_2-c_3\\ \left|\begin{array}{ccc}a-1& 0& 1\\ 1-b& b-1& 1\\ 0& 1-c& c\end{array}\right|=0\\ c\left(a-1\right)\left(b-1\right)+\left(1-b\right)\left(1-c\right)-\left(1-c\right)\left(a-1\right)=0\\ \frac{c}{1-c}+\frac{1}{1-a}+\frac{1}{1-b}=0\\ \frac{c}{1-c}+1+\frac{1}{1-a}+\frac{1}{1-b}=1\\ \frac{1}{1-c}+\frac{1}{1-a}+\frac{1}{1-b}=1\\ optionDiscorrectanswer.\end{array}$