Question

If $a\hat{i}+\hat{j}+\hat{k}$ , $\hat{i}+b\hat{j}+\hat{k}$ , $\hat{i}+\hat{j}+c\hat{k}$ are coplanar and then $\dfrac{1}{1-a}+\dfrac{1}{1-b}+\dfrac{1}{1-c}$ is
(a) 2
(b) 1
(c) -1
(d) 3

Answer 1

Hint: We have been given three coplanar vectors. According to the condition, the determinant of three coplanar vectors is zero. By using this condition we can develop the expression of $\dfrac{1}{1-a}+\dfrac{1}{1-b}+\dfrac{1}{1-c}$ . The expression for the determinant of 3 vectors $\vec{a},\vec{b},\vec{c}$ as follows,
Determinant = $\left| \begin{matrix}
   {{a}_{x}} & {{a}_{y}} & {{a}_{z}} \\
   {{b}_{x}} & {{b}_{y}} & {{b}_{z}} \\
   {{c}_{x}} & {{c}_{y}} & {{c}_{z}} \\
\end{matrix} \right|$ .

Complete step-by-step answer:
We have been given three vectors equation and we need to find the value of $\dfrac{1}{1-a}+\dfrac{1}{1-b}+\dfrac{1}{1-c}$ .
The three vectors are as follow,
$a\hat{i}+\hat{j}+\hat{k}$, $\hat{i}+b\hat{j}+\hat{k}$ , $\hat{i}+\hat{j}+c\hat{k}$ .
We have been given that these vectors are coplanar.
For the vectors to be coplanar their determinant has to be zero.
From this, we get out condition,
The determinant of three vectors can be written as follows,
Determinant = $\left| \begin{matrix}
   a & 1 & 1 \\
   1 & b & 1 \\
   1 & 1 & c \\
\end{matrix} \right|....................(i)$
Equating the above equation we get,
$\left| \begin{matrix}
   a & 1 & 1 \\
   1 & b & 1 \\
   1 & 1 & c \\
\end{matrix} \right|=0$ .
We need to simplify this using column transformation to find the unknown result.
Performing the column transformation,
By applying ${{C}_{1}}={{C}_{1}}-{{C}_{3}}$ , and ${{C}_{2}}={{C}_{2}}-{{C}_{3}}$ , we get,
$\left| \begin{matrix}
   a-1 & 0 & 1 \\
   0 & b-1 & 1 \\
   1-c & 1-c & c \\
\end{matrix} \right|=0$ .
Now taking $\left( a-1 \right)$ common from the first row, $\left( b-1 \right)$ common from the second row, $\left( 1-c \right)$ common from the third row, we get,
$\left( a-1 \right)\left( b-1 \right)\left( 1-c \right)\left| \begin{matrix}
   1 & 0 & \dfrac{1}{a-1} \\
   0 & 1 & \dfrac{1}{b-1} \\
   1 & 1 & \dfrac{c}{1-c} \\
\end{matrix} \right|=0$
To satisfy this condition we need to make the remaining determinant to zero.
Solving the determinant we get,
$\left[ \left( \dfrac{c}{1-c} \right)-\left( \dfrac{1}{b-1} \right) \right]+\left[ \dfrac{1}{a-1}\left( -1 \right) \right]=0$
Simplifying further we get,
$\dfrac{c}{1-c}=\dfrac{1}{a-1}+\dfrac{1}{b-1}$
Subtracting $\dfrac{1}{1-c}$ from both sides we get,
$\dfrac{c}{1-c}-\dfrac{1}{1-c}=\dfrac{1}{a-1}+\dfrac{1}{b-1}-\dfrac{1}{1-c}$
Taking a negative sign from R.H.S. we get,
$\dfrac{-1\left( 1-c \right)}{1-c}=\dfrac{1}{a-1}+\dfrac{1}{b-1}+\dfrac{1}{c-1}$
$\Rightarrow -1=\dfrac{1}{a-1}+\dfrac{1}{b-1}+\dfrac{1}{c-1}$
Taking the negative sign we get,
$\dfrac{1}{1-a}+\dfrac{1}{1-b}+\dfrac{1}{1-c}=1$ ,
Therefore, the value of the expression we need to find is 1.
Hence, the correct option is (b).

Note: It is the common mistake that the sum is ended at $-1=\dfrac{1}{a-1}+\dfrac{1}{b-1}+\dfrac{1}{c-1}$ which gives the option as (c). Also, we need to understand the columns are vertical and rows are counted as horizontal, it is often miscalculated and the whole matrix calculation can go wrong. We also have to be cautious when taking common from each row we have taken $\left( 1-a \right)$ and $\left( 1-b \right)$ from the first two rows and $\left( c-1 \right)$ the third row.