If the vectors \(a \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}, \hat{\mathbf{i}}+b…
If the vectors \(a \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}, \hat{\mathbf{i}}+b \hat{\mathbf{j}}+\hat{\mathbf{k}}\) and \(\hat{\mathbf{i}}+\hat{\mathbf{j}}+c \hat{\mathbf{k}}\) are coplanar, where \((a, b, c \neq 1\) ), then the value of \(\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}=\)
2
0
-1
1
Solution
For coplanar vectors,
\(\Rightarrow \quad\left|\begin{array}{ccc}
a_x & a_y & a_z \\
b_x & b_y & b_z \\
c_x & c_y & c_z
\end{array}\right|=0\)
Substituting values we have, \(\left|\begin{array}{lll}a & 1 & 1 \\ 1 & b & 1 \\ 1 & 1 & c\end{array}\right|=0\)
\(R_2 \rightarrow R_2-R_1 \text { and } R_3 \rightarrow R_3 \rightarrow R_1\)
\(\Rightarrow\left|\begin{array}{ccc}
a & 1 & 1 \\
1-a & b-1 & 0 \\
1-a & 0 & c-1
\end{array}\right|=0\)
\(\Rightarrow a(b-1)(c-1)-(1-a)(c-1)-(1-a)(b-1)=0\)
\(\Rightarrow\) Dividing by \((1-a)(1-b)(1-c)\), we get
\(\begin{aligned}
\frac{a}{1-a}+\frac{1}{1-b}+\frac{1}{1-c} & =0 \\
\Rightarrow \quad \frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c} & =\frac{1}{1-a}-\frac{a}{1-a}=1
\end{aligned}\)