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If the vectors $\hat{\mathbf{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}, \hat{\mathrm{i}}+\hat{\mathrm{b}}+\hat{\mathrm{k}}$ and $\hat{\mathrm{i}}=\hat{\mathrm{j}}=\mathrm{c} \hat{\mathrm{k}}(\mathrm{a}, \mathrm{b}, \mathrm{c} \neq 1)$
are coplanar, then the value of $\frac{1}{1-\mathrm{a}}+\frac{1}{1-\mathrm{b}}+\frac{1}{1-\mathrm{c}}$ is equal to
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are coplanar, then the value of $\frac{1}{1-\mathrm{a}}+\frac{1}{1-\mathrm{b}}+\frac{1}{1-\mathrm{c}}$ is equal to
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The correct answer is:
1
$a \hat{i}+\hat{j}+\hat{k}, \hat{i}+b \hat{j}+\hat{k}$ and $\hat{i}+\hat{j}+c \hat{k}$ are coplanar.
i.e., $\left|\begin{array}{lll}\mathrm{a} & 1 & 1 \\ 1 & \mathrm{~b} & 1 \\ 1 & 1 & \mathrm{c}\end{array}\right|=0$
$c_{2} \rightarrow c_{2}-c_{3} ; c_{3} \rightarrow c_{3}-c_{1}$
$\left|\begin{array}{ccc}a & 1-a & 1-a \\ 1 & b-1 & 0 \\ 1 & 0 & c-1\end{array}\right|=0$
$\Rightarrow \mathrm{a}(\mathrm{b}-1)(\mathrm{c}-1)-(1-\mathrm{a})(\mathrm{c}-1)-(1-\mathrm{a})(\mathrm{b}-1)=0$
Divide both sides by $(1-a)(1-b)(1-c)$
$\begin{aligned} \Rightarrow & \frac{\mathrm{a}(\mathrm{b}-1)(\mathrm{c}-1)}{(1-\mathrm{a})(1-\mathrm{b})(1-\mathrm{c})}-\frac{(1-\mathrm{a})(\mathrm{c}-1)}{(1-\mathrm{a})(1-\mathrm{b})(1-\mathrm{c})} \\ & \quad-\frac{(1-\mathrm{a})(\mathrm{b}-1)}{(1-\mathrm{a})(1-\mathrm{b})(1-\mathrm{c})}=\frac{0}{(1-\mathrm{a})(1-\mathrm{b})(1-\mathrm{c})} \\ \Rightarrow & \frac{\mathrm{a}}{1-\mathrm{a}}+\frac{1}{1-\mathrm{b}}+\frac{1}{1-\mathrm{c}}=0 \\ \Rightarrow & \frac{1}{1-\mathrm{b}}+\frac{1}{1-\mathrm{c}}=-\frac{\mathrm{a}}{1-\mathrm{a}} \end{aligned}$
Add $\frac{1}{1-\mathrm{a}}$ on both sides.
$\Rightarrow \frac{1}{1-\mathrm{a}}+\frac{1}{1-\mathrm{b}}+\frac{1}{1-\mathrm{c}}=\frac{1}{1-\mathrm{a}}-\frac{\mathrm{a}}{1-\mathrm{a}}=\frac{1-\mathrm{a}}{1-\mathrm{a}}=1 .$
i.e., $\left|\begin{array}{lll}\mathrm{a} & 1 & 1 \\ 1 & \mathrm{~b} & 1 \\ 1 & 1 & \mathrm{c}\end{array}\right|=0$
$c_{2} \rightarrow c_{2}-c_{3} ; c_{3} \rightarrow c_{3}-c_{1}$
$\left|\begin{array}{ccc}a & 1-a & 1-a \\ 1 & b-1 & 0 \\ 1 & 0 & c-1\end{array}\right|=0$
$\Rightarrow \mathrm{a}(\mathrm{b}-1)(\mathrm{c}-1)-(1-\mathrm{a})(\mathrm{c}-1)-(1-\mathrm{a})(\mathrm{b}-1)=0$
Divide both sides by $(1-a)(1-b)(1-c)$
$\begin{aligned} \Rightarrow & \frac{\mathrm{a}(\mathrm{b}-1)(\mathrm{c}-1)}{(1-\mathrm{a})(1-\mathrm{b})(1-\mathrm{c})}-\frac{(1-\mathrm{a})(\mathrm{c}-1)}{(1-\mathrm{a})(1-\mathrm{b})(1-\mathrm{c})} \\ & \quad-\frac{(1-\mathrm{a})(\mathrm{b}-1)}{(1-\mathrm{a})(1-\mathrm{b})(1-\mathrm{c})}=\frac{0}{(1-\mathrm{a})(1-\mathrm{b})(1-\mathrm{c})} \\ \Rightarrow & \frac{\mathrm{a}}{1-\mathrm{a}}+\frac{1}{1-\mathrm{b}}+\frac{1}{1-\mathrm{c}}=0 \\ \Rightarrow & \frac{1}{1-\mathrm{b}}+\frac{1}{1-\mathrm{c}}=-\frac{\mathrm{a}}{1-\mathrm{a}} \end{aligned}$
Add $\frac{1}{1-\mathrm{a}}$ on both sides.
$\Rightarrow \frac{1}{1-\mathrm{a}}+\frac{1}{1-\mathrm{b}}+\frac{1}{1-\mathrm{c}}=\frac{1}{1-\mathrm{a}}-\frac{\mathrm{a}}{1-\mathrm{a}}=\frac{1-\mathrm{a}}{1-\mathrm{a}}=1 .$
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