Vectors 3 Question 22
22. If the vectors $a \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}, \hat{\mathbf{i}}+b \hat{\mathbf{j}}+\hat{\mathbf{k}}$ and $\hat{\mathbf{i}}+\hat{\mathbf{j}}+c \hat{\mathbf{k}}(a \neq b \neq c \neq 1)$ are coplanar, then the value of $\frac{1}{(1-a)}+\frac{1}{(1-b)}+\frac{1}{(1-c)}=\ldots \ldots .$.
(1987, 2M)
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Answer:
Correct Answer: 22. (1)
Solution:
- Since, vectors are coplanar.
$\therefore \quad\left|\begin{array}{lll}a & 1 & 1 \ 1 & b & 1 \ 1 & 1 & c\end{array}\right|=0$
Applying $R _2 \rightarrow R _2-R _1, R _3 \rightarrow R _3-R _1$,
$$ \begin{array}{cc} & \left|\begin{array}{ccc} a & 1 & 1 \\ 1-a & b-1 & 0 \\ 1-a & 0 & c-1 \end{array}\right|=0 \\ \Rightarrow & \left|\begin{array}{ccc} a /(1-a) & 1 /(1-b) & 1 /(1-c) \\ 1 & -1 & 0 \\ 1 & 0 & -1 \end{array}\right|=0 \\ \Rightarrow \quad \frac{a}{1-a}(1)-\frac{1}{1-b}(-1)+\frac{1}{1-c}(1)=0 \\ \Rightarrow \quad \frac{a}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}=0 \\ \Rightarrow \quad \frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-b}+\frac{1}{1-c}=0 \\ \Rightarrow \quad & =1 \end{array} $$
