If the vectors $a \hat{i}+\hat{j}+\hat{k} ; \hat{i}+b \hat{j}+\hat{k} ; \hat{i}+\hat{j}+c \hat{k}(\mathrm{a}…

If the vectors $a \hat{i}+\hat{j}+\hat{k} ; \hat{i}+b \hat{j}+\hat{k} ; \hat{i}+\hat{j}+c \hat{k}(\mathrm{a} \neq \mathrm{b} \neq \mathrm{c} \neq 1)$ are coplanar then $\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}=$
  1. 0
  2. 1
  3. 2
  4. -1

Solution

Given vectors are coplanar $\begin{aligned} & \left|\begin{array}{ccc}a & 1 & 1 \\ 1 & b & 1 \\ 1 & 1 & c\end{array}\right|=0 \\ & \left|\begin{array}{ccc}a & 1 & 1 \\ 1-a & b-1 & 0 \\ 1-a & 0 & c-1\end{array}\right|=0\left[\begin{array}{l}R_2^{\prime} \rightarrow R_2-R_1 ; \\ R_3 \rightarrow R_3-R_1\end{array}\right] \\ & \Rightarrow\left|\begin{array}{ccc}\frac{a}{1-a} & \frac{1}{b-1} & \frac{1}{c-1} \\ 1 & 1 & 0 \\ 1 & 0 & 1\end{array}\right|=0 \\ & \Rightarrow\left|\begin{array}{ccc}\frac{a}{1-a} & \frac{1}{b-1} & \frac{1}{c-1} \\ 1 & 1 & 0 \\ 0 & -1 & 1\end{array}\right|=0\left[\text {Applying } R_3 \rightarrow R_3-R_2\right] \\ & \Rightarrow \frac{a}{1-a}-\left(\frac{1}{b-1}+\frac{1}{c-1}\right)=0 \\ & \Rightarrow \frac{1}{1-a}-1+\frac{1}{1-b}+\frac{1}{1-c}=0 \Rightarrow \frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}=1\end{aligned}$

Asked in: AP EAMCET 2024 (23 May Shift 1)

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