Question

If the vectors $a\overrightarrow{i}+\overrightarrow{j}+\overrightarrow{k},\overrightarrow{i}+b\overrightarrow{j}+\overrightarrow{k},\overrightarrow{i}+\overrightarrow{j}+c\overrightarrow{k}$ (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

Answer: (D)

1

We have,

$\left|\begin{array}{cc}a& 11\\ 1& b1\\ 1& 1c\end{array}\right|=0\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}=......$

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

$\left(a-1\right)\left[\left(b-1\right)c-1+c\right]+1\left[\left(1-b\right)\left(1-c\right)\right]=0$

$c\left(a-1\right)\left(b-1\right)-\left(1-c\right)\left(a-1\right)+\left(1-b\right)\left(1-c\right)=0$

$\frac{c}{1-c}+\frac{1}{1-a}+\frac{1}{1-b}=0$

$\frac{c}{1-c}+1+\frac{1}{1-a}+\frac{1}{1-b}=0+1$

$\frac{c}{1-c}+\frac{1}{1-a}+\frac{1}{1-b}=1$

Then,

Option $D$ is correct answer.