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NIMCET 2021 Question Paper with solutions

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Question : 29 of 120

Marks: +1, -0

If the vector

a i+j+k, i+b j+k, i+j+c k(a, b, c \neq 1)

are coplanar then

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

0
1
2
3

Solution:

For coplanar vector

[\vec{a}, \vec{b}, \vec{c}] = 0

\begin{vmatrix} a & 1 & 1 \\ 1 & b & 1 \\ 1 & 1 & c \end{vmatrix} = 0

R_2 \rightarrow R_2 - R_1

and

R_3 \rightarrow R_3 - R_1

\begin{vmatrix} a & 1 & 1 \\ 1-a & b-1 & 0 \\ 1-a & 0 & c-1 \end{vmatrix} = 0

Expand

R_1

a(b-1)(c-1) - 1(1-a)(c-1)

+ 1(a-1)(b-1) = 0

a(b-1)(c-1) + (a-1)(c-1)

+ (a-1)(b-1) = 0

Divide by

(a-1)(b-1)(c-1)

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

Take - ve sign common

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

Add both side 1

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

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

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