Question

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

• A. $-1$
• B. $0$
• C. $1$
• D. $\frac{1}{3}$

Answer 1

The correct option is C

$1$

Explanation for the correct option :

Find the value of $\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}$,

As the vectors $ai+j+k,i+bj+k,i+j+ck$ are coplanar, so $\left[ai+j+ki+bj+ki+j+ck\right]=0$ and thus:

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

Now, in the determinant perform transformations $C_1\to C_1-C_2$ and $C_2\to C_2-C_3$.

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

Expand the determinant.

$\begin{array}{rcl}\left(a-1\right)\left[c\left(b-1\right)-\left(1-c\right)\right]-0+1\left[\left(1-b\right)\left(1-c\right)-0\right]& =& 0\\ c\left(a-1\right)\left(b-1\right)-\left(a-1\right)\left(1-c\right)+\left(1-b\right)\left(1-c\right)& =& 0\\ c\left(1-a\right)\left(1-b\right)+\left(1-a\right)\left(1-c\right)+\left(1-b\right)\left(1-c\right)& =& 0\end{array}$

Divide both sides by $\left(1-a\right)\left(1-b\right)\left(1-c\right)$.

$\begin{array}{rcl}\frac{c\left(1-a\right)\left(1-b\right)+\left(1-a\right)\left(1-c\right)+\left(1-b\right)\left(1-c\right)}{\left(1-a\right)\left(1-b\right)\left(1-c\right)}& =& \frac{0}{\left(1-a\right)\left(1-b\right)\left(1-c\right)}\\ \frac{c}{1-c}+\frac{1}{1-b}+\frac{1}{1-a}& =& 0\end{array}$

Now add $1$ both sides of the equation.

$\begin{array}{rcl}\frac{c}{1-c}+\frac{1}{1-b}+\frac{1}{1-a}+1& =& 0+1\\ \left(\frac{c}{1-c}+1\right)+\frac{1}{1-b}+\frac{1}{1-a}& =& 1\\ \frac{c+1-c}{1-c}+\frac{1}{1-b}+\frac{1}{1-a}& =& 1\\ \frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}& =& 1\end{array}$

Hence, the correct option is C.