Let a, b, c be three distinct real numbers, none equal to one. If the vectors ai∧+j∧+k∧,i∧+bj∧+k∧ and i∧+j∧+ck∧ are coplanar, then 11−a+11−b+11−c is equal to

Given vectors are Coplanar∴a111b111c=0⇒a−1011−bb−1101−cc=0 C1:C1−C2;C2:C2−C3⇒(a−1)(b−1)(c−1)101a−1−111b−10−1cc−1=0⇒101a−1−111b−10−1cc−1=0 a≠1 ; b≠1; c≠1⇒ cc−1+1b−1+1a−1=0 ⇒11−a+11−b+c1−c+1=1⇒11−a+11−b+11−c=1

Let a, b, c be three distinct real numbers, none equal to one. If the vectors ai∧+j∧+k∧,i∧+bj∧+k∧ and i∧+j∧+ck∧ are coplanar, then 11−a+11−b+11−c is equal to

Given vectors are Coplanar∴a111b111c=0⇒a−1011−bb−1101−cc=0 C1:C1−C2;C2:C2−C3⇒(a−1)(b−1)(c−1)101a−1−111b−10−1cc−1=0⇒101a−1−111b−10−1cc−1=0 a≠1 ; b≠1; c≠1⇒ cc−1+1b−1+1a−1=0 ⇒11−a+11−b+c1−c+1=1⇒11−a+11−b+11−c=1

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

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Detailed Solution

Given vectors are Coplanar
$∴ |\begin{matrix} a & 1 & 1 \\ 1 & b & 1 \\ 1 & 1 & c \end{matrix}| = 0$
$\Rightarrow |\begin{matrix} a - 1 & 0 & 1 \\ 1 - b & b - 1 & 1 \\ 0 & 1 - c & c \end{matrix}| = 0 C_{1} : C_{1} - C_{2}; C_{2} : C_{2} - C_{3}$
$\Rightarrow (a - 1) (b - 1) (c - 1) |\begin{matrix} 1 & 0 & \frac{1}{a - 1} \\ - 1 & 1 & \frac{1}{b - 1} \\ 0 & - 1 & \frac{c}{c - 1} \end{matrix}| = 0$
$\Rightarrow |\begin{matrix} 1 & 0 & \frac{1}{a - 1} \\ - 1 & 1 & \frac{1}{b - 1} \\ 0 & - 1 & \frac{c}{c - 1} \end{matrix}| = 0$ $a \neq 1; b \neq 1; c \neq 1$
$\Rightarrow \frac{c}{c - 1} + \frac{1}{b - 1} + \frac{1}{a - 1} = 0 \begin{array}{l} \Rightarrow \frac{1}{1 - a} + \frac{1}{1 - b} + \frac{c}{1 - c} + 1 = 1 \\ \Rightarrow \frac{1}{1 - a} + \frac{1}{1 - b} + \frac{1}{1 - c} = 1 \end{array}$