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Let $a, b$, cbe three distinct real numbers, none equal to one. If the vectors a \hat{i}+\hat{j}+\hat{k}, \hat{i}+b \hat{j}+\hat{k}and \hat{i}+\hat{j}+c \hat{k} are coplanar, then \frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c} is equal to

Option: 1

1


Option: 2

2


Option: 3

-2


Option: 4

-1


Answers (1)

best_answer

\begin{aligned} & \left|\begin{array}{lll} a & 1 & 1 \\ 1 & b & 1 \\ 1 & 1 & c \end{array}\right|=0 \\ & C_2 \rightarrow C_2-C_1, C_3 \rightarrow C_3-C_1 \end{aligned}

\begin{aligned} & \left|\begin{array}{ccc} a & 1-a & 1-a \\ 1 & b-1 & 0 \\ 1 & 0 & c-1 \end{array}\right|=0 \\ & a(b-1)(c-1)-(1-a)(c-1)+(1-a)(1-b)=0 \\ & a(1-b)(1-c)+(1-a)(1-c)+(1-a)(1-b)=0 \\ & \frac{a}{1-a}+\frac{a}{1-b}+\frac{a}{1-c}=0 \\ & \Rightarrow-1+\frac{1}{1-a}+\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}

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