Let a , b , c be three distinct real numbers, none equal to one. If the vectors a i ^ + j ^ + k ^ , i ^ + b j ^ + k ^ and i ^ + j ^ + c k ^ are coplanar, then 1 1 - a + 1 1 - b + 1 1 - c is equal to [2023]

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Solution

Q.
Let $a, b, c$ be 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 [2023]

1 - 2

2 1

3 - 1

4 2

Ans.
(2)

$Given, vectors a \hat{i} + \hat{j} + \hat{k}, \hat{i} + b \hat{j} + \hat{k} and \hat{i} + \hat{j} + c \hat{k} are coplanar.$

$∴$ $| \begin{matrix} a & 1 & 1 \\ 1 & b & 1 \\ 1 & 1 & c \end{matrix} |$ $= 0$

$C_{2} \to C_{2} - C_{1};$ $C_{3} \to C_{3} - C_{1}$

$\Rightarrow$ $| \begin{matrix} a & 1 - a & 1 - a \\ 1 & b - 1 & 0 \\ 1 & 0 & c - 1 \end{matrix} |$ $= 0$

$Taking (1 - a), (1 - b), (1 - c) common from R_{1}, R_{2} and R_{3},$

$\Rightarrow (1 - a) (1 - b) (1 - c)$ $| \begin{matrix} \frac{a}{1 - a} & 1 & 1 \\ \frac{1}{1 - b} & - 1 & 0 \\ \frac{1}{1 - c} & 0 & - 1 \end{matrix} |$ $= 0$

$Expanding, we get$

$(1 - a) (1 - b) (1 - c) [\frac{a}{1 - a} + \frac{1}{1 - b} + \frac{1}{1 - c}] = 0$

$\Rightarrow \frac{a}{1 - a} + \frac{1}{1 - b} + \frac{1}{1 - c} = 0 \Rightarrow \frac{a}{1 - a} + \frac{1}{1 - b} + \frac{1}{1 - c} + 1 = 1$

$\Rightarrow \frac{a}{1 - a} + 1 + \frac{1}{1 - b} + \frac{1}{1 - c} = 1 \Rightarrow \frac{a + 1 - a}{1 - a} + \frac{1}{1 - b} + \frac{1}{1 - c} = 1$

$\Rightarrow \frac{1}{1 - a} + \frac{1}{1 - b} + \frac{1}{1 - c} = 1$

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